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

Students can cross-reference their work with ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(a) to ensure accuracy.

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

Question 1.
One set of 100 observations has the mean 15 and another set of 150 observations has the mean 16. Find the mean of 250 observations by combining the two sets of given observations.
Solution:
Here, n₁ = 100 ; \(\bar{x}_1\) = 15 and n₂ – 150 ; \(\bar{x}_2\) = 16
∴ combined Mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{100 \times 15+150 \times 16}{100+150}\) = \(\frac{1500+2400}{250}\) = \(\frac{3900}{250}\) = 15.6

Question 2.
The mean age of 40 students is 6 years and the mean age of another group of 60 students is 20 years. Find out the mean age of the 100 students combined together.
Solution:
Here n₁ = 40; \(\bar{x}_1\) = 16 and n₂ = 60; \(\bar{x}_2\) = 20
∴ combined Mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{40 \times 16+60 \times 20}{40+60}\) = \(\frac{640+1200}{100}\) = \(\frac{1840}{100}\) = 18.4
Thus required mean age be 18.4 years.

Question 3.
The mean of marks obtained in an examination by a group of 100 students is found to be 49.46. The mean of the marks obtained in the same examination by another group of 200 students was 52.32. Find the mean of the marks obtained by both the groups of students taken together.
Solution:
Here, n₁ = 100; \(\bar{x}_1\) = 49. 46 and n₂ = 200; \(\bar{x}_2\) = 52.32
Thus marks obtained by first group of students = n₁\(\bar{x}_1\)
and marks obtained by second group of students = n₂\(\bar{x}_2\)
Hence, required mean marks obtained by both groups of students taken together = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{100 \times 49.46+200 \times 52.32}{100+200}\) = \(\frac{4946+10464}{300}\) = 51.366

Question 4.
The number of students in section X A and X B are 30 and 35 respectively. The mean scores of students in the mathematics test are as follows:

X A	X B	X A and X B combined
70	?	62

Find the mean score of XB.
Solution:
Given n₁ = 30; n₂ = 35; \(\bar{x}_1\) = 70; \(\bar{x}\)₁₂ = 62; \(\bar{x}_2\) = ?
We know that \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)
⇒ 62 = \(\frac{30 \times 70+35 \times \bar{x}_2}{30+35}\)
⇒ 4030 – 2100 = 35\(\bar{x}_2\)
⇒ \(\bar{x}_2\) = \(\frac { 1930 }{ 35 }\)
= 55.14

Question 5.
Two samples of sizes 50 and 100 are given. The mean of these samples respectively are 56 and 50. Find the mean of size 150 by combining.
Solution:
Here n₁ = 50; n₂ = 100; \(\bar{x}_1\) = 56; \(\bar{x}_2\) = 50
Then \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)
⇒ \(\bar{x}\)₁₂ = \(\frac{50 \times 56+100 \times 50}{50+100}\)
⇒ \(\bar{x}\)₁₂ = \(\frac{2800+5000}{150}\) = \(\frac{7800}{150}\) = 52

Question 6.
The mean and standard deviation of the marks obtained by the groups of students, consisting of 50 each, are given below. Calculate the mean and standard deviation of the marks obtained by all the 100 students.

Groups	Mean	Standard deviation
1	60	8
2	55	7

Solution:
Here n₁ = n₂ = 50; \(\bar{x}_1\) = 60; \(\bar{x}_2\) = 55; σ₁ = 8; σ₁ = 7
Then combined mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{50 \times 30+50 \times 44}{50+50}\) = \(\frac{3000+2750}{100}\) = \(\frac{5750}{100}\) = 57.5
∴ d₁ = \(\bar{x}\)₁₂ – \(\bar{x}_1\) = 57.5 – 60 = – 2.5 and d₂ = \(\bar{x}\)₁₂ – \(\bar{x}_2\) = 57.5 – 55 = 2.5
Therefore, combined S.D = σ₁₂ =

Question 7.
The mean and standard deviation of distribution of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Solution:
Here n₁ = 100; n₂ = 150; \(\bar{x}_1\) = 50; \(\bar{x}_2\) = 40
σ₁ = 5; σ₂ = 6
∴ Combined Mean \(\bar{x}\)₁₂ = \(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\) = \(\frac{100 \times 50+150 \times 40}{100+150}\) = \(\frac{11000}{250}\) = 44
∴ d₁ = \(\bar{x}\)₁₂ – \(\bar{x}_1\) = 44 – 50 = – 6 and d₂ = \(\bar{x}\)₁₂ – \(\bar{x}_2\) = 44 – 40 = 4
∴ Combined S.D = σ₂ =