OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Ex 28(c)

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.