# GSEB Solutions Class 11 Statistics Chapter 3 Measures of Central Tendency Ex 3.2

Gujarat Board Statistics Class 11 GSEB Solutions Chapter 3 Measures of Central Tendency Ex 3.2 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 3 Measures of Central Tendency Ex 3.2

Question 1.
The mean daily wage paid to 75 skilled workers of a factory was â‚¹ 280 whereas the mean daily wage paid to 125 unskilled workers was â‚¹ 150. Find the mean wage of all the workers.
Here, n1 = No. of skilled workers = 75
xÌ„1 = Mean of daily wages of skilled workers = â‚¹ 280
n2 = No. of unskilled workers =125
xÌ„2 = Mean of daily wages of unskilled workers = â‚¹ 150

Mean of daily wages of workers :
Combined mean xÌ„c = $$\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}$$
Putting n1 = 75, xÌ„1 = 280, n2 = 125 and xÌ„2 = 150 in the formula,
xÌ„c = $$\frac{(75 \times 280)+(125 \times 150)}{75+125}$$
âˆ´ xÌ„c = $$\frac{21000+18750}{200}$$
âˆ´ xÌ„c = $$\frac{39750}{200}$$
âˆ´ xÌ„c = â‚¹ 198.75
Hence, the mean of daily wages of workers = 198.75.

Question 2.
Find the weighted mean of the percentage change in prices from the following data:

Here. weights corresponding to the percentage change in price are given. Therefore, for the calculation of weighted mean the table Isprepared as follows:

Weighted mean of percentage change of price:
xÌ„w = $$\frac{\Sigma w x}{\Sigma w}$$
Putting Î£wx = 3779 and Î£w = 32 in the formula,
xÌ„w = $$\frac{3779}{32}$$
âˆ´ xÌ„w = 118.09%
Hence, weighted mean of percentage change in price = 118.09%.

Question 3.
2 officers, 10 clerks and 3 peons contributed for a staff picnic. The contribution collected per person is shown in the following table:

 Officer Clerk Peon â‚¹ 1000 â‚¹ 500 â‚¹200

Find the mean contribution per person using weighted mean.
For weighted mean, we take the number of staff as weight. calculation is shown in the following table:

Mean of individual contribution:
xÌ„w = $$\frac{\Sigma w x}{\Sigma w}$$
Putting Î£wx = 7600 and Î£w =15 in the formula,
xÌ„w = $$\frac{7600}{15}$$
xÌ„w = â‚¹ 506.67
Hence, mean of individual contribution = â‚¹ 506.67.

Question 4.
The mean marks of a student in 7 theory papers is 62. What should be the mean marks in 3 practical examinations so that his mean marks for the entire examination is 68?
(The marks of each theory paper and practical examination are the same.)
Here, n1 = No. of theory papers = 7
xÌ„1 = Mean of marks of theory papers = 62
n2 = No. of papers for practical exam = 3
xÌ„2 = Mean of marks of practical exam = ?
xÌ„c = Mean of marks of theory and practical = 68
Now, xÌ„c = $$\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}$$
Putting xÌ„c = 68, n1 = 7, xÌ„1 = 62 and n2 = 3 in the formula,
68 = $$\frac{(7 \times 62)+\left(3 \times \bar{x}_{2}\right)}{7+3}$$
68 Ã— 10 = 434 + 3xÌ„2
3xÌ„2 = 680 – 434
3xÌ„2Â = 246
xÌ„2 = $$\frac{246}{3}$$
xÌ„2 = 82 marks
Hence, mean of marks of practical exam = 82 marks.