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.

Answer:

Here, n_{1} = No. of skilled workers = 75

xÌ„_{1} = Mean of daily wages of skilled workers = â‚¹ 280

n_{2} = 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 n_{1} = 75, xÌ„_{1} = 280, n_{2} = 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:

Answer:

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.

Answer:

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.)

Answer:

Here, n_{1} = No. of theory papers = 7

xÌ„_{1} = Mean of marks of theory papers = 62

n_{2} = 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, n_{1} = 7, xÌ„_{1} = 62 and n_{2} = 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„_{2}Â = 246

xÌ„_{2} = \(\frac{246}{3}\)

xÌ„_{2} = 82 marks

Hence, mean of marks of practical exam = 82 marks.