Category: OP Malhotra Solutions

Well-structured Class 11 ISC Maths OP Malhotra Solutions Chapter 31 Moving Average Chapter Test facilitate a deeper understanding of mathematical principles.

S Chand Class 11 ICSE Maths Solutions Chapter 31 Moving Average Chapter Test

Question 1.
The following table gives the recorded monthly sales figures of a certain type of television for the 18-month period commencing 1st January 1989.

Calculate the 6-monthly moving averages and display these and the original figures on the same graph using the same axes for both.
Comment briefly on the purpose of moving average graphs.
Solution:
Calculation of six monthly moving average.

Question 2.
The following table gives the numbers of failures of commercial industries in a country during the years 1975 to 1990 .

Draw a graph illustrating the figures.
Calculate the 4-yearly moving average and plot them on the same graph.
Solution:

Question 3.
The average number, in lakhs, of working days lost in strikes during each year of the period 1981-90 was

1981	1982	1983	1984	1985	1986	1987	1988	1989	1990
1.5	1.8	1.9	2.2	2.6	2.7	2.2	6.4	3.6	5.4

Calculate the 3-yearly moving average and draw the moving average graph.
Solution:

Year	No. of working days lost during strike	3-yearly moving total	3-yearly moving average
1981	1.5
1982	1.8	5.2	1.73
1983	1.9	5.9	1.97
1984	2.2	6.7	2.23
1985	2.6	7.5	2.5
1986	2.7	7.5	2.5
1987	2.2	11.3	3.77
1988	6.4	12.2	4.07
1989	3.6	15.4	5.13
1990	5.4

Question 4.
The profit of a soft drink firm (in thousands of rupees) during each month of the year is as given below :

Month	Jan.	Feb.	Mar.	April	May	June	July	Aug.	Sep.	Oct.	Nov.	Dec.
Profit (in thousands of rupees)	3.6	4.3	4.3	3.4	4.4	5.4	3.4	2.4	3.4	1.8	0.8	1.2

Calculate the 4-monthly moving averages and plot these and the original data on a graph sheet.
Solution:
Calculation of 4 yearly moving average

Question 5.
The quarterly profits of a small scale industry (in thousands of rupees) is as follows :

Year	Quarter 1	Quarter 2	Quarter 3	Quarter 4
2012	39	47	20	56
2013	68	59	66	72
2014	88	60	60	67

Calculate 4-quarterly moving averages. Display these and the original figures graphically on the same graph sheet.
Solution:

Question 6.
The number of road accidents in the city due to rash driving over a period of 3 years, is given in the following table :

Year	Jan. – Mar.	April. – June	July – Sept.	Oct. – Dec.
2010	70	60	45	72
2011	79	56	46	84
2012	90	64	45	82

Calculate four quarterly moving averages and illustrate them and original figures on one graph using the same axes for both.
Solution:
Calculation of quarterly moving average

Effective Class 11 ISC Maths OP Malhotra Solutions Chapter 31 Moving Average Ex 31 can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 31 Moving Average Ex 31

Question 1.
The table shows the number of students in a school getting at least a grade C in mathematics for the years 1994 to 2001.
(i) Represent this data as a time series.

1994	1995	1996	1997	1998	1999	2002	2001
97	118	115	117	121	125	111	125

(ii) Calculate the 3-point moving average and plot it on the same graph.
(iii) Are the school’s maths results improving?
(iv) Explain why this is not a good way to work out whether the school’s results are improving.
Solution:
(i) The table of values is given as under :


(iii) The set of data is smoothed out and is increasing.
(iv) The number is increasing but might be because the school is getting bigger.

Question 2.
The profits of a soft drink firm in thousands of litres during each month of a year were :

Calculate 3-monthly moving averages and illustrate graphically. (ISC 1992 type)
Solution:
Calculation of 3-monthly moving arrange is given as :

Month	Profit	3-monthly moving total	3-monthly moving average
Jan.	1.2	–	–
Feb.	0.8	3.4	1.13
Mar.	1.4	3.8	1.267
Apr.	1.6	5.0	1.67
May	2.0	5.6	1.87
June	2.0	7.6	2.53
July	3.6	10.4	3.47
Aug.	4.8	11.8	3.93
Sep.	3.4	10.0	3.33
Oct.	1.8	5.9	1.97
Nov.	0.7	3.7	1.23
Dec.	1.2	–	–

Question 3.
The number of traffic offences committed in a certain city over a period of 3 years is given in the following table :

	Jan-March	April-June	July-Sept.	Oct-Dec.
1968	74	56	48	69
1969	83	52	49	81
1970	94	60	48	79

Calculate 4-quarterly moving averages and illustrate these and original figures on one graph using the same axis for both. Comment briefly on a local politician’s claim that traffic offences were on the increase.
Solution:

Thus the local politician’s claim that traffic offences were on the increase be true, it is cleared from 4-quarterly moving average centred column.

Question 4.
Find the 4-quarterly moving averages in the following table which gives the quarterly index numbers of coal production (for the years 1936-1938). Also plot on the same graph the quarterly index numbers as well as the 4-quarterly moving average. Comment on the nature of the general trend.

Year	Quarters
	1	2	3	4
1936	93.3	81.7	81.5	89.1
1937	93.8	92.3	86.5	93.7
1938	97.6	82.3	79.0	89.3

Solution:

Question 5.
The annual incomes of a firm were recorded every quarter for 4 years. The results are shown in this table.

	1999	2000	2001	2002
1st quarter	₹ 18,00,000	₹ 20,00,000	₹ 21,00,000	₹ 22,50,000
2nd quarter	₹ 14,50,000	₹ 17,80,000	₹ 19,50,000	₹ 21,00,000
3rd quarter	₹ 13,50,000	₹ 15,00,000	₹ 18,00,000	₹ 19,80,000
4th quarter	₹ 19,00,000	₹ 18,30,000	₹ 19,20,000	₹ 20,50,000

(i) Work out the 4-point moving average for the data.
(ii) Plot the original data and the moving average on the same graph.
(iii) Comment on how the firm’s incomes have changed over the 4-years.
Solution:


The data has been smoothed out so there is a steady increase in Income.

Question 6.
The following table shows the daily sales of milk at a local corner shop for a month.

Sun	Mon	Tue	Wed	Thu	Fri	Sat
12	8	6	9	4	11	15
11	7	7	6	3	15	14
14	9	7	7	5	12	15
11	12	8	7	4	14	19

Make a table showing the moving average using a 7-day span, and draw a graph to show the trend of milk sales over the month.
Solution:

Question 7.
The following table gives the monthly expenditure on a motor car for a period of two years.

Calculate 12-month!y moving average for the two years and display them and the original table on the same graph.
Solution:


Question 8.
A new film was shown at a theatre and ran for six w eeks. The attendances are shown in the table.

	Mon	Tue	Wed	Thu	Fri	Sat
First week	243	268	407	384	348	489
Second week	445	501	623	621	527	684
Third week	602	625	800	763	728	800
Fourth week	800	800	800	800	800	800
Fifth week	721	785	800	800	800	800
Sixth week	647	664	683	642	608	726

(i) Plot a line graph from the above data.
(ii) Calculate the 6-day moving average for the data and plot this on the same graph.
(iii) Comment on the weekly attendance.
Solution:

Question 9.
The table below given details of the electricity generated in million kilowatt hoars for public supply in each quarter of the years 1952 to 1955.

Year	Quarter
	1	2	3	4
1952	8.9	7.1	6.7	9.3
1953	10.1	7.5	7.1	10.5
1954	11.7	7.5	8.3	10.9
1955	12.5	8.3	9.5	11.7

Draw a graph illustrating these figures.
Calculating a set of moving averages using the most suitable number of observations; give reasons of your choice. On the same diagram as before draw a graph showing the moving averages.
Solution:
We shall use 4-monthly moving average as it null eliminate the 12 monthly cycle and leave the general trend of the data.

Question 10.
The number of letters, in hundreds, posted in a certain city on each day of a fornight was as follows:

	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
First week	35	70	36	59	62	60	71
Second week	39	72	38	56	63	71	75

Calculate the 7-day moving averages and display these and the original figures graphically on the same diagram, using the same scale and axes. What is the general trend ?
Solution:

Week days	Week days	No. of letters	7-day moving total	7-day moving average
First	Sun	35
	Mon	70
	Tue	36
	Wed	59	393	56.14
	Thu	62	397	56.71
	Fri	60	399	57
	Sat	71	401	57.28
Second	Sun	39	398	56.85
	Mon	72	399	57
	Tue	38	410	58.57
	Wed	56	414	59.14
	Thu	63
	Fri	71
	Sat	75

From graph and table, we observe that the general trend is that the no. of letters posted goes on increasing every day week after week.

Question 11.
In an influenza epidemic the numbers of cases diagnosed were :

Date(Marks)	1	2	3	4	5	6	7	8	9	10	11	12	13	14
Numbers	2	0	5	12	20	27	46	30	31	18	11	5	0	1

On what days do the mode and upper and lower quartiles occur ?
Calculate 3-day moving averages and display them and the original figures on the same graph.
Solution:
Calculation of 3 yearly moving average

Question 12.

Year	Date rate per thousand Quarter ended
	March	June	September	December
1953	13.9	10.3	8.1	10.6
1954	13.8	9.8	7.8	10.8
1955	14.2	10.1	7.8	10.0

Plot these figures on a graph.
Calculate the 4-quarterly moving averages and plot on the same graph.
Solution:

Question 13.
Registered unemployed (hundreds)

	1957	1958
January	638	596
February	602	548
March	509	491
April	462	462
May	359	365
June	295	325
July	290	308
August	322	328
September	377	377
October	392	380
November	480	474
December	542	536
Average for year	439	432.5

Plot these monthly figures on a graph. Calculate the 12-monthly moving averages and plot these on the same graph.
Solution:

Question 14.
A Ballet Company gave a 6-weeks’ season at a large hall capable of seating 4000 people and the attendances in hundreds, at the evening performances, are recorded in the following table.
Attendance, in hundreds, to nearest hundred

	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
First week	12	13	20	19	17	24
Second week	22	25	31	31	26	34
Third week	30	31	40	38	36	40
Fourth week	40	40	40	40	40	40
Fifth week	38	39	40	40	40	40
Sixth week	32	33	34	32	30	36

Plot a graph of the above time-series and include on the same diagram the graph of 6-day moving averages.
Comment on the weekly cycle on attendances and state, with reasons, if you think, an extension of the season of eight weeks, would have been justified.
Solution:

Question 15.
Production of passenger cars, U.S.A. (tens of thousands)

Year	Quarter
	I	II	III
1927	26	36	24
1928	29	36	36
1929	40	52	43

Calculate the 4-quarterly moving averages and then draw the graphs of the given series and the moving averages. Briefly comment on the general trend.
Solution:

From table and graph it is observed that, the demand of cars was increasing year after year.

Question 16.
The aggregate number, in millions, of working days lost in strikes during each year of the period 1950-60 was

1950	‘51	‘52	‘53	‘54	‘55	‘56	‘57	‘58	‘59	‘60
1.4	1.7	1.8	2.2	2.5	3.8	2.1	8.4	3.5	5.3	3.0

Draw a graph to represent this information. Calculate the 3-yearly moving averages and draw the 3-yearly moving averages graph, using the same axes and scales. What is the main purpose in drawing moving average graph ? Comment on whether the purpose is achieved in this case.
Solution:

Years	Working day lost	3-yearly moving total	3-yearly moving average
1950	1.4
1951	1.7	4.9	1.63
1952	1.8	5.7	1.9
1953	2.2	6.5	2.17
1954	2.5	8.5	2.83
1955	3.8	8.4	2.8
1956	2.1	14.3	4.77
1957	8.4	14.0	4.67
1958	3.5	17.2	5.73
1959	5.3	11.8	3.93
1960	3.0

The main purpose in drawing moving average graph is to find the general trend. The purpose is achieved in this case, because the graph shows that the number of working lost during strike is goes on increasing.

Question 17.
The profits of a soft drink firm in thousands of rupees during each month of a year were

Jan.	Feb.	Mar.	Apr.	May	June	July	Aug.	Sept.	Oct.	Nov.	Dec.
1.2	0.8	1.4	1.6	2.0	2.4	3.6	4.8	3.4	1.8	0.8	1.2

Plot these on a graph.
Calculate 4-monthly moving averages and plot these on the same graph. Comment on the general trend.

The general comment on trend is that the profits go on increasing from Jan. to August and start decreasing from Sept, to December.

Question 18.
Calculate,5-yearIy moving averages for the following data of the commercial and industrial failures in a country from 1982 to 1997.

Year	1982	1983	1984	1985	1988	1989	1990	1991	1992	1993	1996	1997
No. of families	23	26	28	32	12	10	9	13	11	14	3	1

Display the actual and tend values on the same graph using the same axes for both.
Solution:

Year	No. of failures	5-yearly moving total	5-yearly moving average
1982	23
1983	26
1984	28	121	24.2
1985	32	108	21.6
1988	12	91	18.2
1989	10	76	15.2
1990	9	55	11
1991	13	57	11.4
1992	11	50	10
1993	14	42	8.4
1996	3
1997	1

Question 19.
The table given below shows the daily attendance in thousands at a certain exhibition over a period of two weeks :

Week 1	52	48	64	68	52	70	72
Week 2	55	47	61	65	58	75	81

Calculate 7-day moving averages and illustrate these and original information on the same graph using the same scales.
Solution:
Calculation of 7-yearly moving average

Question 20.
The profit of a soft-drink firm (in thousands of rupees) during each month of the year is as given below :

Jan.	Feb.	Mar.	Apr.	May	June	July	Aug.	Sept.	Oct.	Nov.	Dec.
3.6	4.3	4.3	3.4	4.4	5.4	3.4	2.4	3.4	1.8	0.8	1.2

Calculate the 4-monthly moving averages and plot these and the original data on a graph sheet.
Solution:
Calculation of 4 yearly moving average

Parents can use Class 11 ISC Maths OP Malhotra Solutions Chapter 30 Index Numbers Chapter Test to provide additional support to their children.

S Chand Class 11 ICSE Maths Solutions Chapter 30 Index Numbers Chapter Test

Question 1.
Construct the consumer price index number for 1990 , taking 1989 as the base year and using simple average of price relative method for the following data :

Commodities	Price in 1989	Price in 1990
Butter	20	21
Cheese	16	12
Milk	3	3
Egg	2.80	2.80

Solution:
We construct the table of values as under:

using simple average of price relatives method
required index number = P₀₁ = \(\frac{1}{\mathrm{~N}} \Sigma\left(\frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\right)\) = \(\frac{1}{4} \times 380\) = 95

Question 2.
A small industrial concern used three raw materials A, B and C in the manufacturing process, the prices of the materials was as shown below :

Commodity	Price (in ₹) in the year 1995	Price (in ₹) in the year 2005
A	4	5
B	60	57
C	36	42

Using 1995 as the base year, calculate a simple aggregate price index for 2005.
Solution:
We construct the following table as under :

By simple aggregate method, we have
required price index for 2015 = P₀₁ = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\) = \(\frac{104}{100} \times 100\) = 104

Question 3.
Find the consumer price index number for the year 2010 as the base year by using method of weighted aggregates.

Commodity	A	B	C	D	E
Year 2000 price (in ₹) per unit	16	40	0.50	5.12	2.00
Year 2010 price (in ₹) per unit	20	60	0.50	6.25	1.50
weights	40	25	5.00	20.00	10.00

Solution:
We construct the table of values as under:

Thus by weighted aggregate method,
required index number = \(\frac{\Sigma \mathrm{P}_1 w}{\Sigma \mathrm{P}_0 w} \times 100\) = \(\frac{2442.5}{1764.9} \times 100\) = 138.39

Question 4.
The price of six different commodities for years 2009 and year 2011 are as follows :

Commodities	A	B	C	D	E	F
Price in 2009 (₹)	35	80	25	30	80	x
Price in 2011 (₹)	50	y	45	70	120	105

The index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x and y if the total price in 2009 is ₹ 360.
Solution:
We construct the table of values given as under :

Commodities	Price in 2009 p0	Price in 2011 p1
A	35	50
B	80	Y
C	25	45
D	30	70
E	80	120
F	x	105
	Σp0 = 250 + x	Σp1 = 390 + y

Since total price in 2009 = ₹ 360 ⇒ 250 + x = 360 ⇒ x = 110
Using simple aggregate method, index number = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\)
⇒ 125 = \(\frac{390+y}{360} \times 100\)
\(\frac{125 \times 36}{10}\) = 390 + y
⇒ 390 + y = 450
⇒ y = 60

The availability of Class 11 ISC Maths OP Malhotra Solutions Chapter 30 Index Numbers Ex 30(b) encourages students to tackle difficult exercises.

S Chand Class 11 ICSE Maths Solutions Chapter 30 Index Numbers Ex 30(b)

Question 1.
Explain briefly, what is meant by a “Weighted average.”
Calculate a cost of living index from the following table of prices and weights.

	weight	Price index
Food	35	108.5
Rent	9	102.6
Clothes	10	97.0
Fuel	7	100.9
Miscellaneous	39	103.7

Solution:
Construct the table of values as under :

	Weight w	Price Index I	Iw
Food	35	108.5	3797.5
Rent	9	102.6	923.4
Clothes	10	108.5	970
Fuel	7	102.6	706.3
Miscellaneous	39	108.5	4044.3
	Σw = 100		ΣIw = 10441.5

Thus, by weighted average of pure relative method,
Cost of living index = \(\frac{\Sigma \mathrm{I} w}{\Sigma w}\) = \(\frac{10441.5}{100}\) = 104.415

Question 2.
Taking 1975 as the base year with an index number 100 , calculate an index number for 1985 based on weighted average of price relatives derived from the table given below :

Commodity	A	B	C	D
Weight	20	30	10	40
Price per unit in 1975	10	20	5	40
Price per unit in 1985	30	35	10	80

Solution:
We construct table of values is given as under:

Then by weighted average of price relatives,
Price Index = \(\frac{\Sigma w x}{\Sigma w}\) = \(\frac{21250}{100}\) = 212.50

Question 3.
Calculate the index number for the year 1979 with 1970 as base from the following data using weighted average of price relatives.

Commodity	Weights	Price in ₹
		1970	1979
A	22	2.50	6.20
B	48	3.30	4.40
C	17	6.25	12.75
D	13	0.65	0.90

Solution:
We construct the table as follows:

Thus by weighted average of price relative method
required index number = \(\frac{\Sigma I w}{\Sigma w}\) = \(\frac{17123.846}{100}\) = 171.24

Question 4.
Construct a composite index number, as a weighted mean from the following data :

Index number	122	145	101	98	137	116
Weight	7	2	4	1	6	5

Solution:
We know that composite index number is the average of index number for different groups of variables. Construct a table of values is given as under :

Index Number I	Weight w	Iw
122	7	854
145	2	290
101	4	404
98	1	98
137	6	822
116	5	580
	Σw = 25	ΣIw = 3048

Required price index = \(\frac{\Sigma \mathrm{I} w}{\Sigma w}\) = \(\frac{3048}{25}\) = 121.92

Question 5.
Construct a composite index number from the following index numbers and weights :

Index Number	127	142	186	172	115
Weight	5	4	3	6	8

Solution:
Construct a table of values is given as under :

Index Number I	Weight w	Iw
127	5	635
142	4	568
186	3	558
172	6	1032
115	8	920
	Σw = 26	ΣIw = 3713

Required price Index = \(\frac{\Sigma I w}{\Sigma w}\) = \(\frac{3713}{26}\) = 142.81

Question 6.
A small industrial concern used three raw materials A, B and C in its manufacturing process. The price, in £ pe kg, of these materials are shown below :

	1957	1967
A	4	5
B	60	57
C	36	42

Using 1957 as the base year, calculate for 1967.
(i) a simple aggregate price index.
(ii) price relatives for the three materials and hence a simple average of relatives index. Does either index suffer from any disadvantage ? If the number of kg’s of A, B and C used per year are 30,5 and 10 respectively, calculate a weighted aggregate price index for 1967 using 1957 as the base year.
Solution:
We construct the table as follows :

(i) By simple aggregate method, we have
required price index for 2002 = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\) = \(\frac{104}{100} \times 100\) = 104

(ii) By simple average of price relative method, we have
required price index for 2002 = \(\frac{\Sigma\left(\frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\right)}{\mathrm{N}}\) = \(\frac{336.67}{3}\) = 112.22
Thus by weighted aggregate method, we have
p₀₁ = \(\frac{\Sigma \mathrm{P}_1 w}{\Sigma \mathrm{P}_0 w} \times 100\) = \(\frac{855}{780} \times 100\) ≃ 109.62

Hence by weighted average of price relative, we have
required price index for 2002 = \(\frac{\Sigma I w}{\Sigma w}\) = \(\frac{5391.7}{45}\) = 119.81
The first two indices in (i) and (ii)
suffer the disadvantage that weight are not used and these values do not reflect the true changes in the cost of production. Since 4th index number 119.81 > 100.
Hence cost of production has gone up.

Question 7.
A manufacturer uses 4 raw materials A, B, C, D in the production of a certain commodity. Masses of raw materials used in manufacturing are in the ratio 2 : 3 : 4 : 1. The prices, in ₹, of the materials per kilogram in the years 1978,1980 are given in the following table :

	A	B	C	D
1978	8	12	6	18
1980	9.50	13	7.50	20

Calculate the index number for the total cost of the raw materials used for the manufacture of the commodity in 1980, using 1978 as the base year.
If the commodity is solid for ₹ 5.75 in 1978, calculate the selling price in 1980, on the assumption that selling prices are directly proportional to the cost of raw material.
Solution:

Then by weighted aggregate method, we have
Required index number = \(\frac{\Sigma \mathrm{P}_1 w}{\Sigma \mathrm{P}_0 w} \times 100\) = \(\frac{108}{94} \times 100\) = 104.89
given selling price of commodity in 1978 = ₹ 5.75
∴ required selling price of commodity in 1980 = \(\frac{\Sigma P_1 w}{\Sigma P_0 w} \times 5.75\) = \(\frac{108 \times 5.75}{94}\) = 6.61

Question 8.
The table shows the averages prices of coffee, sugar and milk in 1979 and 1980 , and the weights used to calculate the cost of making a cup of coffee.

	Cost in 1979 (per kg) ₹ (p0)	Cost in 1979 (per kg) ₹ (p1)	Weights (w)
Sugar	3	7	3
Milk	3	3.50	4
Coffee	90	120	2

Calculate, correct to one decimal place, the index number for the cost of a cup of coffee in 1980 using :
(i) weighted price relatives,
(ii) weighted aggregates
taking the index number for 1979 as 100 in each case
Solution:
We construct the table as follows :

(i) By weighted price relative method, we have
required index number for 2010 = \(\frac{\Sigma I w}{\Sigma w} \) = \(\frac{1433.34}{9}\) = 159.3

(ii) By weighted aggregated method, we have
required index no. = \(\frac{\Sigma P_1 w}{\Sigma P_0 w} \times 100\) = \(\frac{275}{201} \times 100\) = 136.8

Question 9.
An enquiry into the budget of the middle class families in a city in England gave the following information :

Expenses on	Food 35%	Rent 15%	Clothing 20%	Fuel 10%	Misc. 20%
Prices (1928)	£ 150	£ 30	£ 75	£ 25	£ 40
Prices (1928)	£ 145	£ 30	£ 65	£ 23	£ 45

What changes in cost of living figures of 1928 as compared with that of 1929 are seen ?
Solution:
We construct table of values is as under :

Then by weight average of price relative method, we have required index no. = \(\frac{\Sigma w x}{\Sigma w}\) = \(\frac{9786.85}{100}\) = 97.8685
Thus living in 1929 was more cheaper as compared to living in 1928.

Question 10.
Calculate the cost of living index number from the following group data :

Group	Weights	Group Index No.
Food	47	247
Fuel and lighting	7	293
Clothing	8	289
House rent	13	100
Miscellaneous	14	236

Solution:
Construct table of values is given as under :

Group	Weights w	Group Index No. I	Iw
Food	47	247	11609
Fuel and lighting	7	293	2051
Clothing	8	289	2312
House rent	13	100	1300
Miscellaneous	14	236	3304
	Σw = 89		ΣIw = 20576

∴ cost of living Index = \(\frac{\Sigma I w}{\Sigma w}\) = \(\frac{20576}{89}\) = 231.19

Question 11.
The following commodities have the given price indices relative to a base of 100. The weights are also given.

	Relative index	Weight
Butter	181	4
Bread	116	12
Tea	110	3
Bacon	152	7

Calculate the new index for this set of commodities.
Solution:
We construct the table as under :

Commodity	Relative Index I	Weight w	Iw
Butter	181	4	724
Bread	116	12	1392
Tea	110	3	330
Bacon	152	7	1064
		Σw = 26	ΣIw = 3510

Thus by weighted average of pure relative method, we have
required Index number = \(\frac{\Sigma \mathrm{I} w}{\Sigma w}\) = \(\frac{3510}{26}\) = 135

Question 12.
Calculate as index number for the second year, taking the first year as base, taking into account the prices of the four commodities (in ₹ per kg) and the weights given here under :

	A	B	C	D
I year	30	28	36	28
II year	42	35	45	42
weight	24	14	6	25

Solution:
We construct table of values is given as under :

Then by weighted aggregate method,
Index no. = \(\frac{\Sigma \mathrm{P}_1 w}{\Sigma \mathrm{P}_0 w} \times 100\) = \(\frac{2818}{2028} \times 100\) = 138.95

Question 13.
Construct the consumer price index number for 1988 on basis of 1998 from the following data :

Commodity	A	B	C	D	E
weights	40	25	5	20	10
Prices(₹ per unit) 1988	16.00	40.00	0.50	5.12	2.00
Prices(₹ per unit) 1998	20.00	60.00	0.50	6.25	1.50

Solution:
We construct table of values is given as under :

Then by weighted average method of price relatives, Index No. =\(\frac{\Sigma w x}{\Sigma w}\) = \(\frac{12441.4}{100}\) = 124.414

Question 14.
Calculate the index number for the year 2006 with 1996 as the base year by the weighted average of price relatives method from the following data.

Commodity	A	B	C	D	E
weight	40	25	5	20	10
Prices(₹ per unit) 1996	32.00	80.00	1.00	10.24	4.00
Prices(₹ per unit) 2006	40.00	120.00	1.00	15.36	3.00

Solution:

Then by weighted average method of price relative, we have
Index number = \(\frac{\Sigma w x}{\Sigma w}\) = \(\frac{13000}{100}\) = 130

Question 15.
Calculate the cost of living index for the following data :

Solution:
We construct the table as follows :

Thus required cost of living index = \(\frac{\Sigma \mathrm{I} w}{\Sigma w}\) = \(\frac{17122.9}{100}\) = 171.229

Question 16.
Find the consumer price index number for 1991 on the base of 1990 from the following data, using the method of weighted relatives :

Item	Quantity	Price in 1990 (₹)	Price in 1991 (₹)
A	20 units	200	320
B	14 units	400	420
C	15 units	100	120
D	18 units	40	60
E	10 units	20	28

Solution:
We construct the table of values as under :

By weighted average of price relative, we have
P01 = Price Index or index number = \(\frac{\Sigma \mathrm{I} w}{\Sigma w}\) = \(\frac{10570}{77}\) = 137.27

Question 17.
From the following data compose price index by applying weighted average of price relatives method using arithmetic means :

Thus by weighted average of price relative method
required price index = \(\frac{\Sigma \mathrm{I} w}{\Sigma w}\) = \(\frac{85000}{440}\) ≃ 193.18

Question 18.
The following table shows the prices per unit in 1980 and 1984 with weights of the commodities A, B, C, D :

Commodity	weights	Price in units in 1980	Price in units in 1984
A	20	25	30
B	25	20	30
C	15	50	70
D	40	5	10

Taking 1980 as base year with an index number 100 , calculate the index number of 1984 based on weighted average of price relatives.
Solution:
We construct the following table as under :

Thus by using weighted average of price relative
required index number = \(\frac{\Sigma \mathrm{I} w}{\Sigma w}\) = \(\frac{16250}{100}\) =162.50

Question 19.
The price quotations of four different commodities for 2001 to 2009 are as given below. Calculate the index number for 2009 with 2001 as the base year by using weighted average of price relative method.

Commodity	weight	Price (in ₹)
		2009	2001
A	10	9.00	4.00
B	49	43.40	5.00
C	36	9.00	6.00
D	4	3.60	2.00

Solution:
We construct the table of values is as under :

Then weighted average method of price relative,
Index number = \(\frac{\Sigma w x}{\Sigma w}\) = \(\frac{12682}{99}\) = 128.10

Regular engagement with Class 11 ISC Maths OP Malhotra Solutions Chapter 30 Index Numbers Ex 30(a) can boost students’ confidence in the subject.

S Chand Class 11 ICSE Maths Solutions Chapter 30 Index Numbers Ex 30(a)

Question 1.

Commodities	A	B	C	D	E
1993 prices (in ₹)	50	40	10	5	2
1995 prices (in ₹)	80	60	20	10	6

Solution:
We construct the table of values is given as under :

Commodities	Price in 1993 P0	Price in 1995 P1
A	50	80
B	40	60
C	10	20
D	5	10
E	2	6
	ΣP0 = 107	ΣP1 = 176

Using simple aggregate method,
P₀₁ = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\) = \(\frac{176}{107}\) × 100 = 164.486

Question 2.

Commodities	A	B	C	D	E	F
1990 prices (in ₹)	40	60	20	50	80	100
1998 prices (in ₹)	50	60	30	70	90	110

Solution:
We construct the table is as under:

Commodities	Prices P0	Prices P1
A	40	50
B	60	60
C	20	30
D	50	70
E	80	90
F	100	110
	ΣP0 = 350	ΣP1 = 410

∴ required price index number = P₀₁
= \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\) = \(\frac{410}{350} \times 100\) = 117.143

Question 3.

Commodities	A	B	C	D
Price in 1997	90	40	90	30
Price in 1998	95	60	110	35

Solution:
We construct the table is as under:

Using simple aggregate method
Price index = P₀₁ = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\) = \(\frac{300}{250} \times 100\) = 120

Question 4.
Using 2005 as base year, the index numbers for the price of a commodity in 2006 and 2007 are 118 and 125. Calculate the index numbers for 2005 and 2007 if 2006 is taken as the base year.
Solution:
Let prices in year 2005, 2006 and 2007 are P₀, P₁ and PP₂ respectively.
Given \(\frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\) = 118 …(1)
and \(\frac{\mathrm{P}_2}{\mathrm{P}_0} \times 100\) = 125 …(2)
Thus index number for 2005 with 2006 as base year = \(\frac{P_0}{P_1} \times 100\) = \(\frac{100}{118} \times 100\) = 84.745
Index number for 2007 with 2006 as base year

Question 5.
Compute a price index for the following by using price relative method.

Commodities	A	B	C	D	E
price in 1991  (in ₹)	20	40	60	80	100
price in 1992 (in ₹)	70	45	70	90	105

Solution:
We construct the table as under :

Question 6.

Commodities	cement	timber	steel	bricks
price in 1969  (in ₹)	5	9.5	35	12
price in 1970 (in ₹)	8	14.3	42	24

Solution:
We construct the table as under:

using price ralative method, price index = \(\frac{\Sigma\left(\frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\right)}{n}\) = \(\frac{630.53}{4}\) = 157.6325

Question 7.
The index number for the following data for the year 2008 , taking 2004 as base year was found to be 116. The simple aggregate method was used for calculation. Find the numerical value of x and y if the sum of the prices in the year 2008 is ₹ 203.

Commodity	Price in (₹) in the year 2004	Price (in ₹) in the year 2008
A	20	25
B	10	30
C	30	15
D	25	45
E	X	35
F	50	y

Solution:
We construct the following table given as under:

Since sum of prices in the year 2011 = ΣP₁ = 150 + y
⇒ 203 = 150 + y
⇒ y = 53
Thus Index number for year 2011 = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\)
⇒ \(116=\left(\frac{150+y}{135+x}\right) \times 100\)
⇒ 116 = \(\frac{203 \times 100}{135+x}\)
⇒ 135 + x = \(\frac{203 \times 100}{116}=175\)
⇒ x = 175 – 135 = 40

Question 8.
Construct index numbers by the simple average of relative method for 1990 and 1991 with 1989 as the base year.

Commodity Price (in ₹)per unit	A	B	C	D	E
1989	100	40	30	10	20
1990	120	45	35	12	22
1991	150	60	45	15	23

Solution:
We construct the table of values is under:

Question 9.
Construct the index number for 1991 taking 1990 as the base year from the following data by simple average of price relative method.

Commodities	A	B	C	D	E
price in 1990 (in ₹)	100	80	160	220	40
price in 1991 (in ₹)	140	120	180	240	40

Solution:
We construct the table as under :

Then by simple average of price relative method,
price index = \(\Sigma \frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\) = \(\frac{611.591}{5}\) = 122.3182

Question 10.
Construct index number from the following data for 1991 and 1992 taking 1990 as base by using the method of simple average of price relatives :

Group	Price in 1990 (in ₹)	Price in 1991 (in ₹)	Price in 1992 (in ₹)
A	20.00	24.00	22.00
B	1.25	1.50	1.00
C	5.00	8.00	8.00
D	2.00	2.25	2.12

Solution:
We construct a table of values is given as under:

By method of simple average of price relatives, we have
price Index for 1991 = \(\frac{\Sigma \frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100}{n}\) = \(\frac{512.5}{4}\) = 128.125
and price index for 1992 = \(\frac{\Sigma \frac{\mathrm{P}_2}{\mathrm{P}_0} \times 100}{n}\) = \(\frac{456}{4}\) = 114

Question 11.
The following data relate to the price of rice in different years.

Year	1988	1989	1990	1991	1992	1993	1994	1995	1996	1997
Price (in₹)	6	7	7	8	10	14	12	13	14	15

Find out price relatives
(i) taking 1988 as base;
(ii) 1992 as base ;
(iii) taking average of 1988,1989 and 1990 as base.
Solution:
We construct table of values is given as under:

(iii) base year value P₀ = \(\frac{6+7+7}{3}\) = \(\frac{20}{3}\)

Question 12.
Compute a price index for the following by (i) simple aggregate and (ii) average of price relative method.

Commodity	A	B	C	D	E	F
price in 1994 (₹)	20	30	10	25	40	50
price in 1999 (₹)	25	30	15	35	45	55

Solution:
We construct the table of values is given as under :

(i) By simple aggregate method, price index = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\) = \(\frac{205}{175} \times 100\) = 117.143

(ii) By average of price method, we have, price index = \(\frac{\Sigma\left(\frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\right)}{n}\) = \(\frac{737.5}{6}\) = 122.92

Question 13.
Construct an index for 1998 taking 1997 as base by Average of Relatives.

Commodity	A	B	C	D	E
Price in 1997	5	4	8	11	2
Price in 1998	7	6	9	12	2

Solution:
We construct the table of values is given as under :

Then by average of relative method, Price index = \(\frac{\Sigma\left(\frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\right)}{n}\) = \(\frac{611.591}{5}\) = 122.32

Question 14.
Construct the consumer price index number for 1990 taking 1989 as the base year and using simple average of price relative method for the following data :

Commodities	Price in 1989	Price in 1990
Butter	20	21
Cheese	16	12
Milk	3	3
eggs	2.80	2.80

Solution:
We construct the table of values is given as under :

Using simple average of price relatives method
required index number = P₀₁ = \(\frac{1}{N} \Sigma\left(\frac{P_1}{P_0} \times 100\right)\) × 380 = 95

Regular engagement with ISC Class 11 Maths Solutions S Chand Chapter 29 Correlation Analysis Chapter Test can boost students’ confidence in the subject.

S Chand Class 11 ICSE Maths Solutions Chapter 29 Correlation Analysis Chapter Test

Question 1.
Find the coefficient of correlation from the following pairs of observations :
(1,3), (2,2), (3,5), (4,4), (5,6)
Solution:
The table of values is given as under :

\(\bar{X}\) = \(\frac{\Sigma \mathrm{X}}{n}\) = \(\frac{15}{5}\) = 3;
\(\bar{Y}\) = \(\frac{\Sigma \mathrm{Y}}{n}\) = \(\frac{20}{5}\) = 4;
∴ r = \(\frac{\Sigma d \mathrm{X} d \mathrm{Y}}{\sqrt{\Sigma d \mathrm{X}^2} \sqrt{\Sigma d \mathrm{Y}^2}}\) = \(\frac{8}{\sqrt{10} \sqrt{10}}\) = \(\frac{4}{5}\) = 0.8

Question 2.
Find the Karl Pearson’s coefficient of correlation between x and y for the following data :

x	16	18	21	20	22	26	27	15
y	22	25	24	26	25	30	33	14

Solution:
Let Assumed mean for series X be 20 i.e. A = 20 and for series Y be 25 i.e. B = 25, Here n = 8
We construct the table of values is as under :

X	Y	u = X – 20	v = Y – 25	uv	u2	v2
16	22	-4	-3	12	16	9
18	25	-2	0	0	4	0
21	24	1	-1	-1	1	1
20	26	0	1	0	0	1
22	25	2	0	0	4	0
26	30	6	5	30	36	25
27	33	7	8	56	49	64
15	14	-5	11	55	25	121
		Σu = 5	Σv = -1	uv = 152	Σu2 = 135	Σv2 = 221

Question 3.
From the following data, calculate the Karl Pearson’s coefficient of correlation, it being given that \(\bar{y}\) = 8.

x	6	2	10	4	8
Y	?	11	5	8	7

Solution:
Given \(\bar{X}\) = 6 and \(\bar{Y}\) = 8 = \(\frac{9+11+f+8+7}{5}\) ⇒ 40 = 35 + f ⇒ f = 5

By formula, we have
r = \(\frac{\Sigma(\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})}{\sqrt{\Sigma(\mathrm{X}-\overline{\mathrm{X}})^2} \sqrt{\Sigma(\mathrm{Y}-\overline{\mathrm{Y}})^2}}\) = \(\frac{-26}{\sqrt{40} \sqrt{20}}\) = \(\frac{-26}{\sqrt{800}}\) = \(\frac{-26}{20 \sqrt{2}}\) = -0.9192

Question 4.
Calculate Karl Pearson’s coefficient between the values of x and y if Σx = 18, Σx² = 90, n = 10, Σy = 25, Σy² = 120, Σxy = 65.
Solution:
Given Σx = 18, Σx² = 90 ; n = 10 ; Σy = 25 ; Σy² = 120 and Σxy = 65
Karl Pearson coeff. of correlation =

Question 5.
A psychologist selected a random sample of 22 students. He grouped them in 11 pairs so that the students in each pair have nearly equal scores in an intelligence test. In each pair, one student was taught by method A and the other by method B and examined after the course. The marks obtained by them after the course are as follows :

Pairs	1	2	3	4	5	6	7	8	9	10	11
Method A	24	29	19	14	30	19	27	30	20	28	11
Method B	37	35	16	26	23	27	19	20	16	11	21

Calculate Spearman’s Rank correlation.
Solution:
We construct the table of values is as under :

Pairs	Method A	Rank	Method B	Rank	d = R1 – R2	d2
1	24	6	37	1	5	25
2	29	3	35	2	1	1
3	19	8.5	16	9.5	-1	1
4	14	10	26	4	6	36
5	30	1.5	23	5	-3.5	12.25
6	19	8.5	27	3	5.5	30.25
7	27	5	19	8	-3	9
8	30	1.5	20	7	-5.5	30.25
9	20	7	16	9.5	-2.5	6.25
10	28	4	11	11	-7	49
11	11	11	21	6	5	25
						Σd2 = 225

Question 6.
In a contest the competitors were awarded marks out of 20 by two judges. The scores of the 10 competitors are given below. Calculate spearman’s rank correlation.

Pairs	A	B	C	D	E	F	G	H	I	J
Judge A	2	11	11	18	6	5	8	16	13	15
Judge B	6	11	16	9	14	20	4	3	13	17

Solution:
The table of values is given as under :

Peer review of ISC Class 11 Maths Solutions S Chand Chapter 29 Correlation Analysis Ex 29(b) can encourage collaborative learning.

S Chand Class 11 ICSE Maths Solutions Chapter 29 Correlation Analysis Ex 29(b)

Find Rank Correlation Coefficient by Spearman’s formula in the following questions.
TYPE 1. [Based on the formula R = 1 – \(\frac{6 \Sigma D^2}{n\left(n^2-1\right)}\)]

Question 1.
The marks obtained by nine students in Physics and Mathematics are given below :

Physics	48	60	72	62	56	40	39	52	30
Mathematics	62	78	65	70	38	54	60	32	31

Calculate spearman’s coefficient correlation and interpret the result.
Solution:
We construct table of values is given as under:

X	Y	Ranks in X R1	Ranks in Y R2	D = R1 – R2	D2
48	62	4	6	-2	4
60	78	7	9	-2	4
72	65	9	7	2	4
62	70	8	8	0	0
56	38	6	3	3	9
40	54	3	4	-1	1
39	60	2	5	-3	9
52	32	5	2	3	9
30	31	1	1	0	0
					ΣD2 = 40

∴ Spearman coeff. of correlation R = 1 – \(\frac{6 \Sigma \mathrm{D}^2}{n\left(n^2-1\right)}\) = 1 – \(\frac{6 \times 40}{9\left(9^2-1\right)}\) = 1 – \(\frac{6 \times 40}{9 \times 80}\) = \(\frac { 2 }{ 3 }\) = 0.667
This shows that there is a positive moderate relationship between marks in Physics and Mathematics.

Question 2.
In a skating competition the judges gave the five competitors the following marks :

Competitors	A	B	C	D	E
1st judge	5.7	5.8	5.9	5.6	5.5
2nd judge	5.6	5.7	6.0	5.5	5.8

Calculate a coefficient of rank correlation.
Solution:
The table of values is given as under :

1st judge	2nd judge	R1	R2	D2 = (R1 – R2)2
5.7	5.6	3	2	1
5.8	5.7	4	3	1
5.9	6.0	5	5	0
5.6	5.5	2	1	1
5.5	5.8	1	4	9
				ΣD2 = 12

∴ Spearman coeff. of correlation R = 1 – \(\frac{6 \Sigma \mathrm{D}^2}{n^3-n}\) ⇒ R = 1 – \(\frac{6 \times 12}{5^3-5}\) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\) = 0.4

Question 3.
The marks in history and mathematics of twelve students in a public examination are given in the table below :

Student	A	B	C	D	E	F	G	H	I	J	K	L
History	69	36	39	71	67	76	40	20	85	65	55	34
Mathematics	33	52	71	25	79	22	83	81	24	35	46	64

Calculate a coefficient of correlation by ranks. What deduction can be made from the result?
Solution:
The table of values is given as under :

Marks in history I	Marks in Maths II	R1	R2	D2 = (R1 – R2)2
69	33	9	4	25
36	52	3	7	16
39	71	4	9	25
71	25	10	3	49
67	79	8	10	4
76	22	11	1	100
40	83	5	12	49
20	81	1	11	100
85	24	12	2	100
65	35	7	5	4
55	46	6	6	0
34	64	2	8	36
				ΣD2 = 508

∴ ρ = 1 – \(\frac{6 \Sigma \mathrm{D}^2}{n^3-n}\) = 1 – \(\frac{6 \times 508}{12^3-12}\) = 1 – \(\frac{3048}{1716}\) = – 0.776
So there is a negative correlation of higher degree which means that a good student of history is a very bad student of mathematics and vice-versa.

Question 4.
The marks given to five competitors by three different judges were as follows :

Competitors	A	B	C	D	E
Judge X	10	9	4	3	1
Judge Y	8	4	9	3	6
Judge Z	6	8	10	2	4

The result was decided by the average mark of the two judges whose marks showed the best correlation. Calculate :
(i) the coefficient of correlation by ranks for each pair of judges ;
(ii) the final order of the competitors.
Solution:
The table of values is given as under :

Judge X	Judge Y	Judge Z	R1	R2	R3	D212 = (R1 – R2)2	D223 = (R2 – R3)2	D213 = (R1 – R3)2
10	8	6	5	4	3	1	1	4
9	4	8	4	2	4	4	4	0
4	9	10	3	5	5	4	0	4
3	3	2	2	1	1	1	0	1
1	6	4	1	2	2	4	1	1
						ΣD212 = 14	ΣD223 = 6	ΣD213 = 10

Here value of R₂₃ is maximum. Thus the best correlation was between the marks of Judges Y and Z.
Thus the marks obtained by competitors A, B, C, D, E are 14, 12, 19, 5 and 10. Hence the final order of the competitors are C, A, B, E and D.

Question 5.
The coefficient of rank correlation between the marks in Statistics and Mathematics obtained by a certain group of students is \(\frac{2}{3}\) and the sum of the squares of the differences in ranks is 55. Find the number of students in the group.
Solution:
Let n be the required no. of students in the group.

TYPE 2. (using correlation factor)
Question 6.
The final positions of twelve clubs in a football league and the average attendances at their home matches were as follows :

Club	A	B	C	D	E	F	G	H	I	J	K	L
Position	1	2	3	4	5	6	7	8	9	10	11	12
Attendance(thousands)	27	30	18	25	32	12	19	11	32	12	12	13

Calculate a coefficient of correlation by ranks and comment on your result.
What other factors do you think might affect the number of spectators apart from the positions of the clubs in the league?
Solution:
We construct the rank table as follows :

Club	Rank R1	Attendance(thousands)	Rank R2	d = R1 – R2	d2
A	1	27	4	-3	9
B	2	30	3	-1	1
C	3	18	7	-4	16
D	4	25	5	-1	1
E	5	32	1.5	3.5	12.25
F	6	12	10	-4	16
G	7	19	6	1	1
H	8	11	12	-4	16
I	9	32	1.5	7.5	56.25
J	10	12	10	0	0
K	11	12	10	1	1
L	12	15	8	4	16
					Σd2= 145.5

There are 2 ties in Ranks R₂ out of which one tie of 2 items and other tie of 3 items.

Since R = +0.4825 which indicates a moderate positive relation. Apart from position of clubs in the league, many other factors like days of week, whether, accessibility of stadium affect the number of spectators.

Question 7.
The competitors in a beauty contest were awarded marks out of 20 by three judges with the following results :

Competitors	A	B	C	D	E	F	G	H	I	J
Judge X	2	11	11	18	6	5	8	16	13	15
Judge Y	6	11	16	9	14	20	4	3	13	17
Judge Z	13	9	18	16	3	5	6	20	11	15

Determine the rank correlation coefficient in order to decide which two judges have the nearest approach to common taste in beauty. State the final order of the competitors if the result of the competition was decided by the average mark of the two judges having the nearest approach to the common taste in beauty.
Solution:
The table of values is given as under :

Thus R_ZX is Maximum. So we conclude that the pair of first and third judges has the nearest approach to beauty.
Thus the total scores of competitors A, B, C, D, E, F G, H, I and J are :
15, 20, 29, 34, 9, 10, 14, 36, 24, 30 .
Thus the final order of crmpetitors are :
H, D, J, C, I, B, A, G, F and E.

Question 8.
(i) Ten recruits were subjected to a selection test to ascertain their suitability for a certain course of training. At the end of the training, they were given a proficiency test.
The marks, secured by recruits in the selection test x and proficiency test (y) are given below :

Recruits	A	B	C	D	E	F	G	H	I	J
Test (x)	10	16	12	17	13	16	24	14	22	20
Test (y)	30	42	45	46	33	34	40	35	39	45

Calculate a coefficient of correlation by rank and comment on your result.
(ii) Describe the correlation you would expect to find between :
(a) the ages and weights of children under six year old,
(b) the width of a river at different points and the distance of these points from the sea,
(c) the amount of oxygen in the air at a place and its height above sea level.
Note. Use Spearman’s formula in (i).
Solution:

Test x	Text y	Ranks in x R1	Ranks in y R2	D2 = (R1 – R2)2
10	30	1	1	0
15	42	5	7	4
12	45	2	8.5	42.25
17	46	7	10	9
13	33	3	2	1
16	34	6	3	9
24	40	10	6	16
14	35	4	4	0
22	39	9	5	16
20	45	8	8.5	0.25
				ΣD2 = 97.5

which shows that there is a substantial relationship between X and Y.
(ii) (a) As age increases, weight of children are also increases so there is a direct relationship between age and weight of children.
(b) There is an inverse correlation between them.
(c) There is no correlation as amount of oxygen in the air at a place is not only depends upon the height but also on the factors like pollution etc.

Question 9.
The following table gives the record of the heights of eight athletes and the measurements of their long jump and high jump to the nearest cm :

	Height	Long Jump	High Jump
A	158	324	175
B	165	365	185
C	162	380	180
D	170	400	184
E	175	350	200
F	163	350	172
G	178	425	188
H	164	375	180

Calculate the coefficient of rank correlation between height and long jump and between height and high jump. Comment on the result.
Solution:
We construct the following table as under :

Continuous practice using ISC Class 11 Maths Solutions S Chand Chapter 29 Correlation Analysis Ex 29(a) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 29 Correlation Analysis Ex 29(a)

Question 1.
A physicist is experimenting with the resistance in a circuit she is using. She measures and records the resulting current.

Resistance (ohms)	5	10	15	20	25	30	40
Current (amps)	10	4.9	3.2	2.4	1.9	1.7	1.0

(i) Draw a scatter graph of her results.
(ii) Estimate the current for a resistance of 40 ohms.
(iii) Estimate the resistance for a current of 7.5 amps.
Solution:
(i) Plot the points (5, 10), (10, 4.9), (15, 3.2), (20, 2.4), (25, 1.9), (30, 1.7) and (50, 1.0) on graph paper.

(ii) Clearly from scatter diagram, the corresponding current for a resistance of 40 ohms be 1.3 amps.
(iii) Clearly from scatter diagram, the corresponding value of resistance for current of 7.5 amps be 6.5 ohms.

Question 2.
In a small survey the heights of eight boys were measured and their shoe sizes were recorded.

Height (cm)	172	182	164	190	167	169	175	185
Shoe size	8 1/2	10 1/2	7	13	8 1/2	8	10	12

Draw a scatter graphs and use it to find out whether there is a relationship between these sets of data.
Solution:

The plotted points are approximately lie along a straight line suggesting that the shoe size of a boy is related to his height.
TYPE 2. (Based on first formula : \(r=\frac{\Sigma d_x d_y}{\sqrt{\left(\Sigma d_x^2\right)\left(\sum d_y^2\right)}}\), where d_x = x – \(\bar{x}\), d_y = y – \(\bar{y}\))

Question 3.
Calculate Karl Pearson’s coefficient of correlation between the values of X and Y for the following data. Comment on the value of r.

X	1	2	3	4	5
Y	7	6	5	4	3

Solution:
We construct the table of values is as under:

X	Y	X – X̄ X̄ = 3	Y = ȳ ȳ = 5	(X – X̄ ) (Y – ȳ)	(X – X̄)2	(Y – ȳ)2
1	7	-2	2	-4	4	4
2	6	-1	1	-1	1	1
3	5	0	0	0	0	0
4	4	1	-1	1	1	1
5	3	2	-2	-4	4	4
		Σ(X – X̄) = 0		Σ (X – X̄ ) (Y – ȳ) = – 10	Σ(X – X̄)2 = 10	Σ(Y – ȳ)2 = 10

Here \(\bar{X}\) = \(\frac{1+2+3+4+5}{5}\) = \(\frac{15}{5}\) = 3
and \(\bar{Y}\) = \(\frac{7+6+5+4+3}{5}\) = \(\frac{25}{5}\) = 5
Thus coefficient of correlation

since r = – 1, which shows perfect negative correlation between X and Y.

Question 4.

X	1	2	3	4	5	6	7	8	9
Y	12	11	13	15	14	17	16	19	18

Solution:
We construct the table of values is given as under :

X	Y	dX = X – X̄	dy = Y – ȳ	d2x	d2y	dxdy
1	12	-4	-3	16	9	12
2	11	-3	-4	9	16	12
3	13	-2	-2	4	4	4
4	15	-1	0	1	0	0
5	14	0	-1	0	1	0
6	17	1	2	1	4	2
7	16	2	1	4	1	2
8	19	3	4	9	16	12
9	18	4	3	16	9	12
		ΣX = 45	ΣY = 135	Σd2x = 60	Σd2y = 60	Σdxdy = 56

Here \(\bar{X}\) = \(\frac{\Sigma X}{n}\) = \(\frac{45}{9}\) = 5
and \(\bar{Y}\) = \(\frac{\Sigma Y}{n}\) = \(\frac{135}{9}\) = 15
Karl Pearson’s coeff. of correlation \(r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}\) = \(\frac{56}{\sqrt{60} \sqrt{60}}=\frac{56}{60}\) = \(\frac{14}{15}\) = 0.933
which shows that their is a high positive correlation between X and Y.

Question 5.

	X series	Y series
Number of pairs of observation	15	15
Arithmetic mean	25	18
Standard deviation	3.01	3.03
Sum of the squares of deviation from the mean	136	138
Sum of the product of the deviations of x and y	–
series from their respective means	122

Solution:
Given sum of the squares of deviation from the mean of series \(\mathrm{X}=d_{\mathrm{X}}^2=\Sigma(\mathrm{X}-\overline{\mathrm{X}})=136\)
\(d_{\mathrm{Y}}^2=\Sigma(\mathrm{Y}-\overline{\mathrm{Y}})^2=138\)
\(d_{\mathrm{X}} d_{\mathrm{Y}}=\Sigma(\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})=122\)
∴ \(r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}=\frac{122}{\sqrt{136} \sqrt{138}}=0.89\)
So there is high positive correlation between X and Y.

Question 6.
Calculate the Pearson’s coefficient of correlation between the ages of husband and wife.

Age of husband	35	34	40	43	56	20	38
Age of wife	32	30	31	32	53	20	33

Solution:
We construct the table of values is given as under :

Age of husband x	Age of wife y	dx = x – x̄ x̄ = 38	dy = y – ȳ ȳ = 33	dxdy	dx2	dy2
35	32	-3	-1	3	9	1
34	30	-4	-3	12	16	9
40	31	2	-2	-4	4	4
43	32	5	-1	-5	25	1
56	53	18	20	360	324	400
20	20	-18	-13	234	324	169
38	33	0	0	0	0	0
Σx = 266	Σy = 231			Σdxdy = 600	Σdx2 = 702	Σdy2 = 584

\(\bar{x}\) = \(\frac{\Sigma x}{n}\) = \(\frac{266}{7}\) = 38
and \(\bar{y}\) = \(\frac{\Sigma y}{n}\) = \(\frac{231}{7}\) = 33
\(r=\frac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2} \sqrt{\Sigma d_y^2}}=\frac{600}{\sqrt{702} \sqrt{584}}=0.937\)

Question 7.
Given r = 0.8, Σxy = 60, σ_y = 2.5 and Σx² = 90, find the number of items. x and y are deviations from their respective mean.
Solution:
Given r = 0.8 ; Σxy = Σ(x – \(\bar{x}\))(y – \(\bar{y}\))=60 ; σ_y = 2.5 ; Σx² = Σ(x – \(\bar{x}\))² = 90

Calculate Karl Pearson’s coefficient of correlation between the values of x and y for the following data.

Question 8.
(1,2), (2,4), (3,8), (4,7), (5,10), (6,5), (7,14), (8,16), (9,2), (10,20)
Solution:
We construct the table of values is given as under :

X	Y	XY	X2	Y2
1	2	2	1	4
2	4	8	4	16
3	8	24	9	64
4	7	28	16	49
5	10	50	25	100
6	5	30	36	25
7	14	98	49	196
8	16	128	64	256
9	2	18	81	4
10	20	200	100	400
ΣX = 55	ΣY = 88	ΣXY = 586	ΣX2 = 385	ΣY2 = 1114

Question 9.

X	-3	-2	-1	0	1	2	3
Y	9	4	1	0	1	4	9

Solution:
We construct table of values is given as under:

X	Y	XY	X2	Y2
-3	9	-27	9	81
-2	4	-8	4	16
-1	1	-1	1	1
0	0	0	0	0
1	1	1	1	1
2	4	8	4	16
3	9	27	9	81
ΣX = 0	ΣY = 28	ΣXY = 0	ΣX2 = 28	ΣY2 = 196

Question 10.
n = 50, Σx = 75, Σy = 80, Σx² = 150, Σy² = 140, Σxy = 120.
Solution:
Given n = 50 ; Σx = 75 ; Σy = 80 ; Σx² = 150 ; Σy² = 140 and Σxy = 120

Question 11.
n = 10, Σx = 55, Σy = 40, Σx² = 385, Σy² = 192 and Σ(x + y)² = 947.
Solution:
Given n = 10 ; Σx = 55 ; Σy = 40 ; Σx² = 385, Σy² = 192
and Σ(x + y)² = 947
⇒ 948 = Σ(x² + y² + 2xy)
⇒ 947 = Σx² + Σy² + 2Σxy
⇒ 947 = 385 + 192 + 2Σxy
⇒ 947 = 577 + 2Σxy
⇒ 2Σxy = 370
⇒ Σxy = 185

Where u = X – A or \(\frac{\mathbf{X}-\mathbf{A}}{h}\), v = Y – B or \(\frac{\mathbf{Y}-\mathbf{B}}{k}\), A and B being assumed means.

Question 12.

X	16	18	21	20	22	26	27	15
Y	22	25	24	26	25	30	33	14

Solution:
Let Assumed mean for serics X be 20 i.e. A = 20 and for serics Y be 25 i.e. B = 25, Here n = 8 We construct the table of values is an under:

X	Y	u = X – 20	V = Y – 25	uv	u2	v2
16	22	-4	-3	12	16	9
18	25	-2	0	0	4	0
21	24	1	-1	-1	1	1
20	26	0	1	0	0	1
22	25	2	0	0	4	0
26	30	6	5	30	36	25
27	33	7	8	56	49	64
15	14	-5	-11	55	25	121
		Σu = 5	Σv = -1	uv = 152	Σu2 = 135	Σv2 = 221

Thus using formula, we have

Question 13.

X	1	2	4	5	7	8	10
Y	2	6	8	10	14	16	20

Solution:

X	Y	u = X – A A = 5	u = Y – B B = 10	uv	u2	v2
1	2	-4	-8	32	16	64
2	6	-3	-4	12	9	16
4	8	-1	-2	2	1	4
5	10	0	0	0	0	0
7	14	2	4	8	4	16
8	16	3	6	18	9	36
10	20	5	10	50	25	100
		Σu = 2	Σv = 6	Σuv = 122	Σu2 = 64	Σv2 = 236

So there is a positive and perfect correlation between X an Y.

Question 14.
Calculate Karl Pearson’s correlation coefficient between the marks in English and Hindi obtained by 10 students.

Marks in English	10	25	13	25	22	11	12	25	21	20
Marks in Hindi	12	22	16	15	18	18	17	23	24	17

Solution:
We construct the table of values is given as under :

X	Y	u = X – A A = 20	v = Y – 17	uv	u2	v2
10	12	-10	-5	50	100	25
25	22	+5	5	25	25	25
13	16	-7	-1	7	49	1
25	15	+5	-2	-10	25	4
22	18	2	1	2	4	1
11	18	-9	1	-9	81	1
12	17	-8	0	0	64	0
25	23	5	6	30	25	36
21	24	1	7	7	1	49
20	17	0	0	0	0	0
		Σu = -16	Σv = 12	Σuv = 102	Σu2 = 374	Σv2 = 142

Question 15.
Show that the coefficient of correlation ρ between two variables x and y is given by \(\rho=\frac{\sigma_x^2+\sigma_y^2–\sigma_{x-y}^2}{2 \sigma_y \sigma_x}\) where \(\sigma_x^2, \sigma_y^2\) and \(\sigma_{x-y}^2\) are the variances of x, y and x-y respectively.
Solution:

Question 16.
A computer expert while calculating correlation coefficient between X and Y from 25 pairs of observations obtained the following results :
n = 25, ΣX = 125, ΣX² = 650, ΣY = 100, ΣY² = 460, ΣXY = 508
It was, however, later discovered at the time of checking that he had copied down two pairs as while the correct values were Obtain the correct value of correlation coefficient.
Solution:
Given n = 25, ΣX = 125 ; ΣX² = 650 ; ΣY² = 460 ; ΣY = 100 ; ΣXY = 508
Corrected ΣX = Given ΣX – (Sum of incorrect values) + (Sum of correct values) = 125 – (6 + 8) + (8 + 6) = 125
Corrected ΣX² = Given ΣX² – (6² + 8²) + (8² + 6²) = 650 – 100 + 100 = 650
Corrected ΣXY = Given ΣXY – (6 × 14 + 8 × 6) + (8 × 12 + 6 × 8) = 508 – (84 + 48) + (96 + 48) = 520
Corrected ΣY = 100 – (14 + 6) + (12 + 8) = 100
Corrected ΣY² = 460 – (14² + 6²) +(12² + 8²) = 436

Question 17.
A computer obtained the data: n = 30, Σx = 120, Σy = 90, Σx² = 600, Σy² = 250 and Σxy = 56. Later on, it was found that pairs are wrong while the correct values are . Find the correct values of ρ(X, Y).
Solution:
Given n = 30; Σx = 120; Σy = 90; Σx² = 600 ; Σy² = 250 and Σxy = 356
Corrected Σx = Given Σx – (Sum of incorrect values) + (Sum of correct values) = 120 – (8 + 12) + (8 + 10) = 118
Corrected Σx² = 600 – (8² + 12²) + (8² + 10²) = 556
Corrected Σy = 90 – (10 + 7) + (12 + 8) = 93
Corrected Σy² = 250 – (10² + 7²) + (12² + 8²) = 309
Corrected Σxy = 356 – (8 × 10 + 12 × 7) + (8 × 12 + 10 × 8) = 368

Question 18.
Show that Pearson’s coefficient of correlation lies between -1 and +1 , i.e., -1 ≤ r ≤ 1 or | r | ≤ 1.
Solution:

Regular engagement with ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Chapter Test can boost students’ confidence in the subject.

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Chapter Test

Question 1.
The means of two sets of sizes 40 and 60 respectively are 15 and 16 and the standard deviations are 3 and 4. Obtain the mean and standard deviation of the composite set of 100 items when the two sets are pooled together.
Solution:

Question 2.
Find the median of the following values :
7 cm, 9 cm, 10 cm, 12 cm, 15 cm, 18 cm, 20 cm.
Solution:
Given data is already in ascending order :
Here number of observations = n = 7 (odd)
∴ Median = size of \(\left(\frac{n+1}{2}\right)\) th item = size of \(\left(\frac{7+1}{2}\right)\)th item = size of 4th item = 12
Thus median value of 12 cm.

Question 3.
The marks obtained by 12 students out of 50 are as under :
25, 24, 23, 32, 40, 27, 30, 25, 20, 15, 16, 45 Find the median marks.
Solution:
Arranging the given data in ascending order : 15, 16, 20, 23, 24, 25, 25, 27, 30, 32, 40, 45
Here no. of observations = n = 12 (even)

Question 4.
Compute Q₃, D₆ and P₇₀ for the following data :
28, 17, 12, 25, 26, 19, 13, 27, 21, 16
Solution:
Arranging the given data in ascending order, we have
12, 13, 16, 17, 19, 21, 25, 26, 27, 28
Here no. of observations = n = 10 (even)
∴ Q₃ = 3\(\left(\frac{n+1}{4}\right)\) th value = 3\(\left(\frac{10+1}{4}\right)\)th value = 8.25th value
= 8th value + 0.25 (9th value – 8th value) = 26 + 0.25 (27 – 26) = 26.25

D₆ = 6\(\left(\frac{n+1}{10}\right)\)th value = 6\(\left(\frac{10+1}{100}\right)\)th value =6.6th value
= 6th value + 0.6 (7th value – 6th value) = 21 + 0.6 (25 – 21) = 21 + 0.6 × 4 = 21 + 2.4 = 23.4

P₇₀ = 70\(\left(\frac{n+1}{100}\right)\)th value = 70\(\left(\frac{10+1}{100}\right)\)th value = 7.7th value = 7th value + 0.7 (8th value – 7th value) = 25 + 0.7(26 – 25) = 25.7

Question 5.
Obtain the median for the following data and describe the information conveyed by it.

Number of students absent	5	6	7	8	9	10	11	12	13	15	18	20
Number of days	1	5	11	14	16	13	10	70	4	1	1	1

Solution:
We construct the table of values is given as under :

∴ Median = \(\left(\frac{N+1}{2}\right)\)th item = \(\left(\frac{147+1}{2}\right)\)th item = 74th item = 12
Thus shows that for first half, the no. of days 12 or less than 12 students remained absent, while on the remaining half of the no. of days 12 or more than 12 students remained absent.

Question 6.
Calculate the median for the following distribution:

x	1	2	6	9	11	8	5	4
f	5	7	9	11	13	15	17	19

Solution:
We construct the table of values is given as under:

x	1	2	4	5	6	8	9	11
f	5	7	19	17	9	15	11	13	Σf = N = 96
c.f	5	12	31	48	57	72	83	96

∴ Median = \(\left(\frac{\mathrm{N}+1}{2}\right)\)th value = \(\left(\frac{97}{2}\right)\)th value = 48.5 value
= \(\frac { 1 }{ 2 }\)(48th + 49th)obs = \(\frac { 1 }{ 2 }\)(5 + 6) = 5.5

Question 7.
Calculate the median, first quartile and third decile for the following data:

Weekly income (in ₹)	58	59	60	61	62	63	64	65	66
No. of workers	2	3	6	15	10	5	4	3	2

Solution:
We construct the table of values is given as under :

Weekly Income (in ₹)	58	59	60	61	62	63	64	65	66
No. of workers (f)	2	3	6	15	10	5	4	3	2
c.f	2	5	11	26	36	41	45	48	50

Question 8.
Draw an ogive for the following distribution :

Duration (in sec)	0-30	30-60	60-90	90-120	120-150	150-180	180-210
No. of cells	25	75	50	25	150	50	25

Hence, find Q₁, Q₃, D₉, P₄ and P₄₀ from it.
Solution:
Less than type cumulative frequency distribution is given below:

Duration (in s) less than or equal to	30	60	90	120	150	180	210
No. of calls C.f	25	100	150	175	325	375	400

Taking 1 cm along x-axis = 30 sec.
1 cm along y-axis = 50 calls
Plot the points (30,25), (60,100), (90,150), (120,175), (150,325), (180,375), (210,400) and (0,0). Joining all these points by free hand curve gives the required less than type ogive.
More than type cummulative frequency distribution is given below :

Duration (in sec) more than or equal to	0	30	60	90	120	150	180
No. of calls (C.f)	400	375	300	250	225	75	25

Plot the points (0,400), (30,375), (60,300), (90,250), (120,225), (150,75), (180,25), (210,0) and join them by free hand curve give the required more than type ogive.
Thus both ogives intersects at P. From P, draw ⊥ PM on x-axis. The abscissa of point M gives the required median.
Thus, required median = 125.

Question 9.
Calculate the median, Q₃, D₇ and P₇₀ for the following distribution :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of students	3	10	17	7	6	4	2	1

Solution:
The table of values is given as under :

Q₃ : Here \(\frac { 3N }{ 4 }\) = \(\frac{3 \times 50}{4}\) = 37.5, which lies in class 40 – 50.
Here, l = 40 ; f = 6 and c =37 ; i = 10

Median : Here \(\frac { N }{ 2 }\) = \(\frac { 50 }{ 2 }\) = 25, which lies in class 20 – 30.
Here l = 20 ; f = 17 ; c = 13 ; i = 10

Question 10.
Graphically find the mode of the following distribution :

Height (in cm)	30-40	40-50	50-60	60-70	70-80	80-90
No. of students	4	3	8	11	6	2

Solution:
Draw rectangles corresponding to each class interval with heights corresponding to their frequencies. Now join the ends of the opposite rectangles in which rectangle corresponds to modal class be sandwitched between them. Then draw a ⊥ from the point of intersection P meeting x-axis at A.
Then abscissa of point A gives the required mode.
Here modal class be 60 – 70.
∴ required mode = 64

Peer review of ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(d) can encourage collaborative learning.

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(d)

Question 1.
Find the mode of the given distribution by drawing a histogram.

Class Interval	0-10	10-20	20-30	30-40
Frequency	10	45	12	3

Solution:
Draw rectangles corresponding to each class interval with heights corresponds to their frequencies. Here the modal class be 10-20. Now join the ends of the opposite rectangles in which rectangle corresponding to modal class be sandwitched between them. Then draw a ⊥ from point of intersection P to x-axis. Thus x-coordinate of point A gives the required mode.
Clearly Mode = 15.15

Question 2.

Marks	1-5	6-10	11-15	16-20	21-25
f	7	10	16	32	24

Solution:
Here we convert first of all, discontinuous to continuous distribution.
Here adjustment factor = \(\frac{6-5}{2}\) = 0.5
we subtract 0.5 from each lower limit and add 0.5 to each upper limit of each class interval.

Marks	0.5-5.5	5.5-10.5	10.5-15.5	15.5-20.5	20.5-25.5
f	7	10	16	32	24

Draw rectangles corresponding to each class interval with heights corresponds to their frequencies. Here modal class be 16 – 20. Join the ends of the opposite rectangle in which rectangle corresponding to modal class be sandwitched between them. Then draw a ⊥ from point of intersection P meeting x-axis at A. Then abscissa of P gives the required mode.
∴ Required mode = 18.8

Question 3.

Marks	26-30	31-35	36-40	41-45
f	18	10	5	1

Solution:
We make first of all, discontinuous into continuous distribution by using adjustment factor.
Here adjustment factor = \(\frac{31-30}{2}\) = 0.5. So we subtract 0.5 from lower limit and add 0.5 to upper limit of each class interval.
Draw rectangles corresponding to each class interval with heights corresponds to their frequencies. Here modal class be 25.5 – 30.5.
Join RT and SU, draw a ⊥ from point of intersection P meeting x-axis at A.
Then x-coordinate of point A gives the required mode.
∴ Mode = 28.96

Question 4.

Mid-value	60	64	68	72	76
f	25	8	10	66	4

Solution:
The frequency distribution table is given as under:

Mid-value	60	64	68	72	76
Class Interval	58-62	62-66	66-70	70-74	74-78
Frequency	25	8	10	66	4

Draw rectangles corresponding to each class interval with heights corresponding to their frequencies. Now join the ends of the opposite rectangles in which rectangle corresponds to modal class be sandwitched between them. Then draw a ⊥ from the point of intersection P meeting x-axis at A. Then abscissa of point A gives the required mode. ∴ Mode = 71.90

Question 5.

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Freq.	5	11	19	21	16	10	8	6	3	1

Solution:
Draw rectangles corresponding to each class interval with heights corresponding to their frequencies. Now join the ends of the opposite rectangles in which rectangle corresponds to modal class be sandwitched between them. Then draw a ⊥ from the point of intersection P meeting x-axis at A. Then abscissa of point A gives the required mode.

Thus required mode = 32.8

Question 6.
The daily profits in rupees of 100 shops are distributed as under :

Profits per shop (₹)	0-100	100-200	200-300	300-400	400-500	500-600
No. of shops	12	18	27	20	17	7

Draw a histogram of the data and find the modal value. Verify by direct calculation.
Solution:
Draw rectangles corresponding to each class interval with heights corresponding to their frequencies. Now join the ends of the opposite rectangles in which rectangle corresponds to modal class be sandwitched between them. Then draw a ⊥ from the point of intersection P meeting x-axis at A. Then abscissa of point A gives the required mode.
∴ Mode = 256.25

By Inspection, modal class be 200 – 300.
Here l = 200; f_m = 27 ; f_m-1 = 18 ; f_m+1 = 20 ; i = 100
∴ Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 200 + \(\frac{27-18}{54-18-20} \times 100\) =200 + 56.25 = 256.25

Question 7.
Determine the value of mode of the following distribution graphically and verify the result.

Marks	0-10	10-20	20-30	30-40	40-50	50-60
No. of students	5	12	14	10	8	6

Solution:
Draw rectangles corresponding to each class interval with heights corresponding to their frequencies. Now join the ends of the opposite rectangles in which rectangle corresponds to modal class be sandwitched between them. Then draw a ⊥ from the point of intersection P meeting x-axis at A. Then abscissa of point A gives the required mode.
Mode = 23.3

By inspection, modal class be 20 – 30,
we have, l = 20 ; f_m = 14 ; f_m-1 = 12 ; f_m+1 = 10 ; i = 10
∴ Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = \(20+\frac{14-12}{28-12-10} \times 10\) = 20 + 3.33 = 23.33

Question 8.
(i) In an asymmetrical distribution mean is 58 and median is 61 . Calculate mode.
(ii) If mode in a tolerably asymmetrical distribution is 12 and median is 16 , what would be the most probable mean?
(iii) Find the median if mean is 40 and mode is 36.
Solution:
(i) We know that mode = 3 median – 2 man
∴ Mode = 3 × 61 – 2 × 58 = 183 – 116 = 67

(ii) Given mode = 12 ; median = 16
since mode = 3 median – 2 mean
⇒ 12 = 3 × 16 – 2 mean
⇒ 2 mean = 48 – 12 = 36
⇒ mean = 18

(iii) Given mean = 40 ; mode = 36
We know that, Mode = 3 median – 2 mean
⇒ 36 = 3 median – 2 × 40
⇒ 3 median = 36 + 80 = 116
⇒ median = \(\frac{116}{3}\) = 38.67

Continuous practice using ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(c) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(c)

Question 1.
Find the mode of the following data :
(i) 3, 4, 7, 11, 4, 3, 4, 5, 6, 4, 1, 2, 4, 4
(ii) Size of shoes : 4, 4.5, 5, 4.5, 5.5, 5, 6, 4.5, 4, 4.5
(iii) Wages (₹) : 100, 120, 100, 120, 130, 120, 120, 130, 120, 100
(iv) Runs in an innings : 18, 32, 0, 40, 60, 69, 33, 69, 35, 11, 20
Solution:
(i) Arranging the given data in ascending order; we have
1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 11
Here 4 repeated maximum no. of times.
∴ required mode = 4

(ii) Arranging the given data in ascending order : 4, 4, 4.5, 4.5, 4.5, 4.5, 5, 5, 5.5, 6
Here 4.5 repeated maximum no. of times.
∴ required mode = 4.5

(iii) Arranging the given data in ascending order :
100, 100, 100, 120, 120, 120, 120, 120, 130, 130
Here 120 repeated maximum no. of times i.e. 5 times. Thus required mode = 120

(iv) Arranging the given data in ascending order: 0, 11, 18, 20, 32, 33, 35, 40, 60, 69, 69
Here 69 repeated maximum no. of times
∴ required mode = 69

Question 2.
Find the mode from the following data in question 2 to 7.

Marks	10	12	15	22	25	35	45	50	60
Number of students	4	6	10	14	20	19	10	6	3

Solution:
Here maximum frequency be 20 and corresponding observation be 25. Thus required modal marks be 25.

Question 3.

Size	8	10	12	14	16	18	20	22	24
Frequency	3	5	1	7	8	6	4	2	3

Solution:
The table of values is given as under:

∴ Mean = \(\frac{\Sigma f x}{\Sigma f}\) = \(\frac{652}{41}\)
Here, \(\frac{N+1}{2}\) = \(\frac{41+1}{2}\) = 21
Thus M_d = 16
∴ Mode = 3 median – 2 mean = 3 × 16 – 2 × \(\frac{652}{41}\) ≃ 16.19

Question 4.

Class interval	0-10	10-20	20-30	30-40
Frequency	10	45	12	3

Solution:
From given data, maximum frequency be 45 and thus 10 – 20 be the modal class.
Here l = 10; f_m = 45 ; f_m-1 = 10; f_m+1 = 12 ; i = 10
Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 10 + \(\frac{45-10}{90-10-12} \times 10\) = 10 + \(\frac{350}{68}\) = 10 + 5.1470 = 15.147

Question 5.

Marks	0-10	10-20	20-30	30-40	40-50
Number of students	8	16	26	60	38

Solution:
Here maximum frequency be 60 and modal class be 30 – 40.
Here, l = 30; f_m = 60; f_m-1 = 28 ; f_m+1 = 38 ; i = 10
∴ Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 30 + \(\frac{60-28}{120-28-38} \times 10\) =30 + \(\frac{320}{56}\) = 35.71

Question 6.

Marks	Frequency	Marks	Frequency
1-5	7	26-30	18
6-10	10	31-35	10
11-15	16	36-40	5
16-20	32	41-45	1
21-25	24

Solution:
Given class-intervals are discontinuous we make it continuous with the help of adjustment factor.
Here adjustment factor = \(\frac{6-5}{2}\) = 0.5
We subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Here modal class be 15.5 – 20.5
∴ l = 15.5; f_m = 32 ; f_m-1 = 16; f_m+1 = 24; i = 5
Thus,
Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\)
= 15.5 + \(\frac{32-16}{64-16-24} \times 5\)
= 15.5 + \(\frac{80}{24}\) ≃ 18.84

Question 7.

Mid-value	60	64	68	72	76
Frequency	25	8	10	6	4

Solution:
The table of values is given as under:

Mid-value	Class Intervals	Frequency(f)
60	58-62	25
64	62-66	8
68	66-70	10
72	70-74	6
76	74-78	4
		Σf = N = 53

Here Modal class be 58 – 62.
Here
l = 58; f_m = 25 ; f_m+1 = 8 ; i = 4 ; f_m-1 = 0
Thus Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 58 + \(\frac{25-0}{50-8}\) × 4 = 60.38

Question 8.
(i) The mode of the following frequency distribution is 48.6 Find the unknown frequency.

Class	20-25	25-40	40-55	55-70	70-85
Frequency	6	20	44	?	3

(ii) If the frequency of the class 70-85 is 13 instead of 3, then what difference will it make?
Solution:
Let the missing frequency be f.
Here modal class be 40 – 55.
∴ l = 40; f_m = 44 ; f_m-1 = 20 ; f_m+1 = f; i = 15
Thus,
Mode = \( l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m-1}} \times i\)
48.6 = 40 + \(\frac{44-20}{88-20-f} \times 15\)
⇒ 8.6 = \(\frac{360}{68-f}\)
⇒ 584.8 – 8.6f = 360
⇒ 8.6(68 – f) = 360
⇒ 584.8 – 8.6f = 360
⇒ 8.6 f = 224.8
⇒ f = 26.14
If the frequency of class 70 – 85 is 13 instead of 3.
Then value of mode is unaffected as value of mode affected if there are changes in the values of f_m, f_m-1 and f_m+1.

Question 9.
Find the mean, median and mode of the following :
(i) The data 17, 32, 35, 33, 15, 21, 41, 32, 11, 18, 20, 22, 11, 15, 35, 23, 38, 12.
(ii)

Size of item	11	12	13	14	15	16	17
Frequency	8	13	25	37	23	15	9

Solution:
(i) Required Mean = \(\frac{17+32+35+33+15+21+41+32+11+18+20+22+11+15+35+23+38+12}{18}\)
= \(\frac{431}{18}\) = 23.94
Arranging the given data in ascending order ; we have
11, 11, 12, 15, 15, 17, 18, 20, 21, 22, 23, 32, 32, 33, 35, 35, 38, 41
Here no. of observations = n = 18(even)

The observations 11, 15, 32 and 35 repeated equal no. of times i.e. 2 times. Thus, mode is illdefined.

(ii) The table of values is given as under:

∴ Mean = \(\frac{\Sigma f x}{\Sigma f}\) = \(\frac{1825}{130}\) ≃ 14.04
Here, \(\frac{N+1}{2}\) = \(\frac{130+1}{2}\) = 65.5 ∴ Median = 14
Further maximum frequency be 37 and corresponding value of x i.e. 14 gives the required value of mode.

Question 10.
The mode of the following distribution is 240. Find out the missing frequency :

Size	Frequency	Size	Frequency
0-100	140	300-400	?
100-200	230	400-500	150
200-300	270	500-600	140

Solution:
Let the missing frequency be f.
Clearly the maximum frequency be 270 and modal class be 200 – 300.

Here l = 200; f_m = 270 ; f_m-1 = 230 ; f_m+1 = f ; i = 100
Thus, Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) ⇒ 240 = 200 + \(\frac{(270-230)}{540-230-f} \times 100\)
⇒ 40 = \(\frac{4000}{310-f}\) ⇒ 40(310 – f) = 4000 ⇒ 12400 – 40f = 4000 ⇒ 40f = 8400 ⇒f = 210

Question 11.
Find the median, mode, third quartile, 5th decile and 65th percentile from the following :

Value of the item	1	2	3	4	5	6	7	8	9	10
Frequency	25	30	38	45	46	58	32	34	25	22

Solution:
The table of values is given as under:

Here \(\frac{N+1}{2}\) = \(\frac{355+1}{2}\) = 178 ∴ M_d = 5
Here maximum Frequency be 58 and corresponding value of x i.e. 6 be the required value of mode.

Q₃ : Here \(\frac{3(\mathrm{~N}+1)}{4}\) = \(\frac{3 \times 356}{4} \) = 267 and corresponding c.f be 274 and the corresponding value of x gives the required value of Q₃. Thus Q₃ = 7

D₅ : Here, \(\frac{5(\mathrm{~N}+1)}{10}\) = \(\frac{5 \times 356}{10}\) = 178
∴ D₆ = size of 178th item = 5

P₆₅ ; Here, \(\frac{65(\mathrm{~N}+1)}{100}\) = \(\frac{65 \times 356}{100}\) = 231.4
∴ P₆₅ = size of 231.4th item = 6

Question 12.
Find median, mean and modal marks and also determine the limits between which 80% of the students have secured marks.

Marks	No. of students	Marks	No. of students
0-10	15	50-60	80
10-20	25	60-70	70
20-30	52	70-80	10
30-40	56	80-90	5
40-50	78	90-100	2

Solution:
The table of values is given as under:

Then by Step-deviation method, we have
Mean \(\bar{x}\) = A + \(\frac{\Sigma f d^{\prime}}{\mathrm{N}}\) × i = 45 – \(\frac { 15 }{ 393 }\) × 10 = 44.62
Here \(\frac { N }{ 2 }\) = \(\frac { 393 }{ 2 }\) = 196.5 which lies in 40 – 50.
Thus median class be 40 – 50.
Here
l = 40; f = 78; c = 148 ; i = 10
∴ M_d = \(l+\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{f} \times i\) = 40 + \(\frac{196.5-148}{78} \times 10\) = 40 + 6.2195 = 46.22
Clearly modal class be 50 – 60.
Here l = 50 ; f_m = 80 ; f_m-1 = 78 ; f_m+1 = 70 ; i = 10
∴ Mode = \(l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i\) = 50 + \(\frac{80-78}{160-78-70}\) × 10 = 50 + \(\frac { 20 }{ 12 }\) = 51.67
P₁₀: Here \(\frac{10 \times N}{100}\) = \(\frac{10 \times 393}{100}\) = 39.3 which lies in 10 – 20
Here l = 10; c = 15; f = 25; i = 10
\(\mathrm{P}_{90}=l+\frac{\frac{90 \mathrm{~N}}{100}-\mathrm{C}}{f} \times i\) = \(60+\frac{353.7-306}{70} \times 10\) = 66.81
The required limits between which 80% students have secured marks be 19.72 and 66.81.

Effective ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(b) can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(b)

Question 1.
Find the median of the following sets of data:
(i) 2, 3, 5, 7, 9
(ii) 4, 8, 12, 16, 20, 23, 28, 32
(iii) 60, 33, 63, 61, 44, 48, 51
(iv) 13, 22, 25, 8, 11, 19, 17, 31, 16, 10
Solution:
(i) arranging the given data in ascending order; we have 2, 3, 5, 7, 9
Here, no. of observations = n = 5 (odd)
∴ M_d = size of \(\left(\frac{n+1}{2}\right)\)th observation = size of \(\left(\frac{5+1}{2}\right)\)th term = size of 3rd term = 5

(ii) The given data is already in ascending order.
Here no. of observations = n = 8 (even)
∴ Median = \(\frac{\left(\frac{n}{2}\right) \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }}{2}\) = \(\frac{4 \text { th term }+5 \text { th term }}{2}\) = \(\frac{16+20}{2}\) = 18

(iii) Arranging the given data in ascending order; we have, 33, 44, 48, 51, 60, 61, 63
Here no. of observations = n = 7 (odd)
∴ Median = size of \(\left(\frac{n+1}{2}\right)\)th term = size of \(\left(\frac{7+1}{2}\right)\)th term = size of 4th term = 51

(iv) Arranging the given data in ascending order, we get, 8, 10, 11, 13, 16, 17, 19, 22, 25, 31
Here no. of observations = n = 10 (even)
∴ Median = size of =

Question 2.
Find the median of the following data:
41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median?
Solution:
Arranging the given data in ascending order; we get
41, 43, 57, 58, 61, 71, 92, 99, 127,
Here no. of observations = n = 9 (odd)
∴ Median = size of \(\left(\frac{n+1}{2}\right)\)th term = size of \(\left(\frac{9+1}{2}\right)\)th term = size of 5th term = 61
When observation 58 is replaced by 85,
Then data arranged in ascending order is an under : 41, 43, 57, 58, 61, 71, 92, 99, 127,
Then median = size of \(\left(\frac{9+1}{2}\right)\)th term = size of 5th term = 71

Question 3.
Calculate the median of the following incomes.

Wages in ₹	20	21	22	23	24	25	26	27	28
No. of workers	8	10	11	16	20	25	15	9	6

Solution:
The table of values is given as under:

Wages in (₹)	No. of workers	c.f.	Wages in ₹	No. of workers	c.f
20	8	8	25	25	90
21	10	18	26	15	105
22	11	29	27	9	114
23	16	45	28	6	120
24	20	65

Here M_d = \(\left(\frac{n+1}{2}\right)\)th item = size of \(\left(\frac{120+1}{2}\right)\)th item = size of 60.5th term = 24
Thus median income of workers be ₹ 24.

Question 4.
Compute the median of the following distributions:

f	2	3	4	5	6	7
x	3	8	10	12	16	14

Solution:
The table of values is given as under:

x	3	8	10	12	16	14
f	2	3	4	5	6	7	Σf = N = 27
c.f	2	5	9	14	20	27

Here Σf = = 27 ∴ M_d = size of \(\left(\frac{N+1}{2}\right)\)th item = size of 14th item = 12

Question 5.
Marks obtained by 38 students are given below. Calculate the median marks:

Marks	20	90	50	70	60	95
No. of students	4	5	8	10	6	5

Solution:
The table of values is given as under:

Marks	Frequency	c.f
20	4	4
90	5	9
50	8	17
70	10	27
60	6	33
95	5	38
	Σf = N = 38

Here M_d = size of \(\left(\frac{N+1}{2}\right)\)th item = size of \(\left(\frac{38+1}{2}\right)\)th item = size of 19.5th item = 70 [since value 19.5 lies under c.f column in 27]

Question 6.
Find whether the following statements are true or false:
(i) The median of a frequency distribution is the most commonly occurring value.
(ii) The median of a discrete ungrouped frequency distribution containing a number of items is the value of the middle item, the data being arranged in ascending or descending order.
Solution:
(i) False, Mode of a frequency distribution is the most commonly occurring value.
(ii) True, M_d = size of \(\left(\frac{n+1}{2}\right)\)th item

Question 7.
In a school examination it is decided that exactly half the pupils will pass. Name the measure of central tendency that is used.
Solution:
Since by definition of median, equal number of items above and below the median. Thus median’s exactly the middle point of array of given numbers. Since it is given that exactly half the pupils will pass. Thus the measure of central tendency that is used be median.

Question 8.
(1, 2, 3, 6, 8) is a set of five positive integers whose mean is 4 and median is 3. Write down two other sets of five positive integers, each having the same mean and median as this set.
Solution:
For given five observations, mean is 4 and median is 3.
Let the observations 1, 2, 3, 5, 9
∴ Mean = \(\frac{1+2+3+5+9}{5}\) = 4 and M_d = size of \(\left(\frac{5+1}{2}\right)\)th item = size of 3rd item = 3
Let the observations are 1, 2, 3, 4, 10
Here Mean = \(\frac{1+2+3+4+10}{5}\) = 4 and M_d = 3

Question 9.
Find the median from the following distribution:

Marks	10-25	25-40	40-55	55-70	70-85	85-100
Frequency	6	20	44	26	3	1

Solution:
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) 50 (even) which lies in median class 40-55
∴ l = 40; f = 44; C = 26 and i = 15

Here the required median marks are 48.18 marks.

Question 10.

Class Interval	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
Frequency	7	10	13	26	35	22	11	6

Solution:
Here class intervals are discontinuous, we make it continuous by using
adjustment factor = \(\frac { 16-15 }{ 2 }\) = 0.5 which is to be subtracting from lower limit and adding to upper limit of each class interval.
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 130 }{ 2 }\) = 65 which lies in median class 30.5 – 35.5
Thus, l = 30. 5; f = 35; c = 56; i = 5

Question 11.

Value	Frequency	Value	Frequency
Less than 10	4	Less than 50	96
Less than 20	16	Less than 60	112
Less than 30	40	Less than 70	120
Less than 40	76	Less than 80	125

Solution:
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 125 }{ 2 }\) = 62.5 which lies in median class 30 – 40.
Thus, l = 30; f = 36; c = 40; i = 10

Question 12.

Size	Frequency	Size	Frequency
More than 50	0	More than 20	123
More than 40	40	More than 10	165
More than 30	98

Solution:
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 165 }{ 2 }\) = 82.5 which lies in median class 30 – 40.
Thus, l = 30; f = 58; c = 67; i = 10

Question 13.
Find the median and first and third quartile for the following data:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
Solution:
Given data is already in ascending order: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
Here number of observations = n = 11 (odd)
∴ M_d = size of \(\left(\frac{n+1}{2}\right)\)th item = size of \(\left(\frac{11+1}{2}\right)\)th item = size of 6th item = 12
Q₁ = size of \(\left(\frac{n+1}{4}\right)\)th item = size of \(\left(\frac{11+1}{4}\right)\)th item = size of 3rd item = 6
and Q₃ = size of \(3\left(\frac{n+1}{4}\right)\)th item = size of \(\left(\frac{11+1}{4}\right)\)th item = size of 9th item = 18

Question 14.
Compute Q₁, Q₃, D₃, D₆ and D₈ for the following data:
14, 7, 13, 12, 13, 17, 8, 10, 6, 15, 18, 21, 20
Solution:
Arranging the given data in ascending order: 6, 7, 8, 10, 12, 13, 14, 15, 17, 18, 20, 21
Here no. of observations = n = 13 (odd)
Q₁ = size of \(\left(\frac{n+1}{4}\right)\)th item = size of \(\left(\frac{13+1}{4}\right)\)th item = size of 3,5th item = 3rd item + 0.5 (4th item – 3rd item) = 8 + 0.5 (10 – 8) = 8 + 1 = 9

Q₃ = size of \(\left(\frac{n+1}{4}\right)\)th item = size of \(3\left(\frac{13+1}{4}\right)\)th item = size of 10.5th item = 10th item + 0.5 (11th item – 10th item) = 17 + 0.5 (18 – 17) = 17.5

D₃ = size of \(\left(\frac{n+1}{10}\right)\)th item = size of \(3\left(\frac{13+1}{10}\right)\)th item = size of 4.2th item = 4th item + 0.2 (5th item – 4th item) = 10 + 0.2 (12 – 10) = 10 + 0.4 = 10.4

D₆ = size of \(\left(\frac{n+1}{10}\right)\)th item = size of \(6\left(\frac{13+1}{10}\right)\)th item = size of 8.4th item = 8th item + 0.4 (9th item – 8th item) = 14 + 0.4 (15 – 14) = 14.4

D₈ = size of \( 8\left(\frac{n+1}{10}\right)\)th item = size of 11.2th item = 11 th item + 0.2 (12th – 11th) = 18 + 0.2 (20 – 18) = 18.4

Question 15.
Following are the scores of 12 students in a class test of 30 marks:
18, 20, 9, 15, 21, 26, 14, 13, 27, 22, 16, 28 Find D₇ and P₃₃.
Solution:
Arranging the data in ascending order : 9, 13, 14, 15, 16, 18, 20, 21, 22, 26, 27, 28
Here no. of observations = n = 12
∴ D₇ = size of 7\(\left(\frac{n+1}{10}\right)\)th item = size of 7 \(\left(\frac{12+1}{10}\right)\)th item = size of 9.1th item = 9th item + 0.1 (10th – 9th item) = 22 + 0.1 (26 – 22) = 22.4
P₃₃ = size of 33 \(\left(\frac{n+1}{100}\right)\)th item = size of 33 \(\left(\frac{12+1}{100}\right)\)th item = size of 4.29 th item = 4th item + 0.29 (5th item – 4th item) = 15 + 0.29 (16 – 15) = 15.29

Question 16.
Compute Q₁, Q₃, Q₆ and P₄₅ for the following data:

x	18	19	20	21	22	23	24	25	26	27
f	15	18	25	27	40	25	19	16	8	7

Solution:
We construct the following table:

xi	18	19	20	21	22	23	24	25	26	27
fi	15	18	25	27	40	25	19	16	8	7	Σfi = 200
Cumulative freq.	15	33	58	85	125	150	169	185	193	200

Q₁ = \(\left(\frac{n+1}{4}\right)\)th value = \(\left(\frac{200+1}{4}\right)\)th value = 50.25th value
= 50th value + 0.25 (51th value – 50th value) = 20 + 0.25 (20 – 20) = 20 [both 51th and 50th values are same]

Q₃ = 3 \(\left(\frac{n+1}{4}\right)\)th value = 3 \(\left(\frac{200+1}{4}\right)\)th value = 150.75th value = 150th value + 0.75 (151th value – 150th value) = 23 + 0.75 (24 – 23) = 23.75

D₆ = 6 \( \left(\frac{n+1}{10}\right)\)th value = 6 \(\left(\frac{200+1}{10}\right)\)th value = 120.6th value = 120th value + 0.6 (121th value – 120th value) = 22 + 0.6(22 – 22) = 22

P₄₅ = 45 \(\left(\frac{n+1}{100}\right)\)th value = 45 \(\left(\frac{200+1}{100}\right)\)th value = 90.45th value = 90th value + 0.45 (91th value – 90th value) =22 + 0.45 (22 – 22) = 22 (since 91th and 90th values are same)

Question 17.
Following are the different sizes and number of shoes in a shoe shop. Calculate median, first quartile, third quartile, 6th decile and 80th percentile.

Size of shoes	No. of shoes	Size of shoes	No. of shoes
4.5	4	8	40
5	8	8.5	20
5.5	12	9	15
6	15	9.5	24
6.5	20	10	12
7	35	10.5	5
7.5	50	11	3

Solution:

∴ M_d = size of \(\left(\frac{\mathrm{N}+1}{2}\right)\)th item = size of \(\left(\frac{253+1}{2}\right)\)th item = size of 127 th item = 7.5
Q₁ = size of \(\left(\frac{\mathrm{N}+1}{4}\right)\)th item = size of \(\left(\frac{253+1}{4}\right)\)th item = size of 63.5 th item = 7.0
Q₃ = size of 3 \(\left(\frac{\mathrm{N}+1}{4}\right)\)th item = size of 3 \(\left(\frac{253+1}{4}\right)\)th item = size of 190.5 th item = 8.5
D₆ = size of 6 \(\left(\frac{N+1}{10}\right)\)th item = size of 6 \(\left(\frac{253+1}{10}\right)\)th item = size of 152.4 th item = 8
P₈₀ = size of 80 \(\left(\frac{\mathrm{N}+1}{100}\right)\)th item = size of 80\(\left(\frac{253+1}{100}\right)\)th item = 203.2th item = 9.0

Question 18.
Find out first quartile, third quartile and first decile.

Size of Item	f	Size of Item	f
0 – 10	2	40 – 50	34
10 – 20	18	50 – 60	20
20 – 30	30	60 – 70	6
30 – 40	45	70 – 80	3

Solution:
The table of values is given as under:

Question 19.
Calculate the median, 3rd decile and 20th percentile for the following data:

x	0-5	5-10	10-15	15-20	20-25
f	7	18	25	30	20

Solution:
The table of values is given as under:

Question 20.
Find the interquartile range, quartile deviation for the following data:

Age in years	20	30	40	50	60	70	80
No. of members	3	61	132	153	140	51	3

Solution:
The table of values is given as under:

For Q₁ : Q₁ = size of \(\left(\frac{N+1}{4}\right)\)th item = size of \(\left(\frac{543+1}{4}\right)\)th item = size of 136th item = 40

Q₃ = size of 3 \(\left(\frac{\mathrm{N}+1}{4}\right)\)th item = size of 3 \(\left(\frac{543+1}{4}\right)\)th item = size of 408 th item = 60
∴ Interquartile range = Q₃ – Q₁ = 60 – 40 = 20
∴ Q.D = \(\frac{Q_3-Q_1}{2}\) = \(\frac{60-40}{2}\) = 10

Question 21.
Find the interquartile range, semi-interquartile range, and coefficient of quartile deviation from the following frequency distribution.

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90
No. of students	60	45	120	25	90	80	120	60

Solution:
The table of values is given as under:

For Q₁ : \(\frac{N}{4}\) = \(\frac{600}{4}\) = 150, which lies in class interval 30 – 40
Here l = 30; f= 120; c = 105; i = 10
∴ Q₁ = l + \(\frac{\frac{N}{4}-C}{f}\) × i = 30 + \(\frac{150-105}{120}\) × 10 = 30 + 3.75 = 33.75

For Q₃ : \(\frac{3N}{4}\) = \(\frac{3 \times 600}{4}\) = 450 which lies in 70-80
Here l = 70; f= 120; c = 420; i = 10
Q₃ = l + \(\frac{\frac{3 N}{4}-C}{f}\) × i = 70 + \(\frac{450-420}{120}\) × 10 = 72.5
∴ Interquartile range = Q₃ – Q₁ = 72.5 – 33.75 = 38.75
and Semi-interquartile range = \(\frac{Q_3-Q_1}{2}\) = 19.375
and Coeff. of Q.D = \(\frac{Q_3-Q_1}{Q_3+Q_1}\) = \(\frac{72.5-33.75}{72.5+33.75}\) = \(\frac{38.75}{106.25}\) = 0.365