OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Chapter Test

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