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 ₹ 280 whereas the mean daily wage paid to 125 unskilled workers was ₹ 150. Find the mean wage of all the workers.

Answer:

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

x̄_{1} = Mean of daily wages of skilled workers = ₹ 280

n_{2} = No. of unskilled workers =125

x̄_{2} = Mean of daily wages of unskilled workers = ₹ 150

Mean of daily wages of workers :

Combined mean x̄_{c} = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)

Putting n_{1} = 75, x̄_{1} = 280, n_{2} = 125 and x̄_{2} = 150 in the formula,

x̄_{c} = \(\frac{(75 \times 280)+(125 \times 150)}{75+125}\)

∴ x̄_{c} = \(\frac{21000+18750}{200}\)

∴ x̄_{c} = \(\frac{39750}{200}\)

∴ x̄_{c} = ₹ 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:

x̄_{w} = \(\frac{\Sigma w x}{\Sigma w}\)

Putting Σwx = 3779 and Σw = 32 in the formula,

x̄_{w} = \(\frac{3779}{32}\)

∴ x̄_{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 |

₹ 1000 | ₹ 500 | ₹200 |

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:

x̄_{w} = \(\frac{\Sigma w x}{\Sigma w}\)

Putting Σwx = 7600 and Σw =15 in the formula,

x̄_{w} = \(\frac{7600}{15}\)

x̄_{w} = ₹ 506.67

Hence, mean of individual contribution = ₹ 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

x̄_{1} = Mean of marks of theory papers = 62

n_{2} = No. of papers for practical exam = 3

x̄_{2} = Mean of marks of practical exam = ?

x̄_{c} = Mean of marks of theory and practical = 68

Now, x̄_{c} = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)

Putting x̄_{c} = 68, n_{1} = 7, x̄_{1} = 62 and n_{2} = 3 in the formula,

68 = \(\frac{(7 \times 62)+\left(3 \times \bar{x}_{2}\right)}{7+3}\)

68 × 10 = 434 + 3x̄_{2}

3x̄_{2} = 680 – 434

3x̄_{2} = 246

x̄_{2} = \(\frac{246}{3}\)

x̄_{2} = 82 marks

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