# RD Sharma Solutions Class 7 Data Handling Ii Central Values Exercise 23.2

## RD Sharma Solutions Class 7 Chapter 23 Exercise 23.2

#### Exercise 23.2

Q1) A die was thrown 20 times and the following scores were recorded:

5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1

Prepare the frequency table of the scores on the upper face of the die and find the mean score.

Solution:

The frequency table for the given data is as follows:

x:         1          2          3          4          5          6

f:          2          5          1          4          6          2

In order to compute the arithmetic mean, we prepare the following table:

Computation of Arithmetic Mean

 Scores ($x_{i}$) Frequency ($f_{i}$) $x_{i}f_{i}$ 1 2 2 2 5 10 3 1 3 4 4 16 5 6 30 6 2 12 Total $\sum f_{i}$ = 20 $\sum f_{i}x_{i}$ = 73

We have, $\sum f_{i}$ = 20 and $\sum f_{i}x_{i}$ = 73

∴ Mean score = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{73}{20}$ = 3.65.

Q2) The daily wages (in Rs) of 15 workers in a factory are given below:

200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180

Prepare the frequency table and find the mean wage.

Solution:

The frequency table for the given data is as follows:

Wages (xi):                  130      150      180      200

Number of workers (fi):            2          4          6          3

In order to compute the mean wage, we prepare the following table:

Mean wages of the workers

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 130 2 260 150 4 600 180 6 1080 200 3 600 Total $\sum f_{i}$ = N = 15 $\sum f_{i}x_{i}$ = 2540

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{2540}{15}$ = 169.33.

Q3) The following table shows the weights (in kg) of 15 workers in a factory:

Weight (in kg):                        60        63        66        72        75

Numbers of workers:                         4          5          3          1          2

Calculate the mean weight.

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 60 4 240 63 5 315 66 3 198 72 1 72 75 2 150 Total $\sum f_{i}$ = 15 $\sum f_{i}x_{i}$ = 975

∴ Mean Weight = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{975}{15}$ = 65 kg.

Q4) The ages (in years) of 50 students of a class in a school are given below:

Age (in years):                        14        15        16        17        18

Numbers of students:             15        14        10        8          3

Find the mean age.

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 14 15 210 15 14 210 16 10 160 17 8 136 18 3 54 Total $\sum f_{i}$ = 50 $\sum f_{i}x_{i}$ = 770

∴ Mean Weight = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{770}{50}$ = 15.4 yrs.

Q5) Calculate the mean for the following distribution:

x:                     5          6          7          8          9

f:                      4          8          14        11        3

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 5 4 20 6 8 48 7 14 98 8 11 88 9 3 27 Total $\sum f_{i}$ = 40 $\sum f_{i}x_{i}$ = 281

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{281}{40}$ = 7.025.

Q6) Find the mean of the following data:

x:                     19        21        23        25        27        29        31

f :                     13        15        16        18        16        15        13

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 19 13 247 21 15 315 23 16 368 25 18 450 27 16 432 29 15 435 31 13 403 Total $\sum f_{i}$ = N = 106 $\sum f_{i}x_{i}$ = 2650

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{2650}{106}$ = 25.

Q7) The mean of the following data is 20.6. Find the value of p.

x :                    10                    15                    p                      25                    35

f :                     3                      10                    25                    7                      5

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 10 3 30 15 10 150 p 25 25p 25 7 175 35 5 175 Total $\sum f_{i}$ = 50 $\sum f_{i}x_{i}$ = 530 + 25p

We have:

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

=> 20.6 = $\frac{530+25p}{50}$

=> 530 + 25p = 20.6 x 50

=> 25p = 1030 – 530

=> p = $\frac{500}{25}$

=> p = 20

Q8) If the mean of the following data is 15, find p.

x :                    5                      10                    15                    20                    25

f:                      6                      p                      6                      10                    5

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 5 6 30 10 p 10p 15 6 90 20 10 200 25 5 125 Total $\sum f_{i}$ = 27 + p $\sum f_{i}x_{i}$ = 445 + 10p

We have:

$\sum f_{i}$ = 27 + p, $\sum f_{i}x_{i}$ = 445 + 10p

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

=> 15 = $\frac{445+10p}{27+p}$

=> 445 + 10p = 405 + 15p

=> 5p = 445 – 405

=> p = $\frac{40}{5}$

=> p = 8.

Q9) Find the value of p for the following distribution whose mean is 16.6

x:         8          12        15        p          20        25        30

f :         12        16        20        24        16        8          4

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 8 12 96 12 16 192 15 20 300 p 24 24p 20 16 320 25 8 200 30 4 120 Total $\sum f_{i}$ = N = 100 $\sum f_{i}x_{i}$ = 1228 + 24p

We have:

$\sum f_{i}$ = 100, $\sum f_{i}x_{i}$ = 1228 + 24p

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

=> 16.6 = $\frac{1228+24p}{100}$

=> 1228 + 24p = 16.6 x 100

=> 24p = 1660 – 1228

=> p = $\frac{432}{24}$

=> p = 18.

Q10) Find the missing value of p for the following distribution whose mean is 12.58

x :                    5          8          10        12        p          20        25

f:                      2          5          8          22        7          4          2

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 5 2 10 8 5 40 10 8 80 12 22 264 p 7 7p 20 4 80 25 2 50 Total $\sum f_{i}$ = N = 50 $\sum f_{i}x_{i}$ = 524 + 7p

We have:

$\sum f_{i}$ = 50, $\sum f_{i}x_{i}$ = 524 + 7p

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

=> 12.58 = $\frac{524+7p}{50}$

=> 524 + 7p = 12.58 x 50

=> 7p = 629 – 524

=> p = $\frac{105}{7}$

=> p = 15.

Q11) Find the missing frequency (p) for the following distribution whose mean is 7.68

x:                     3                      5                      7                      9                      11                    13

f:                      6                      8                      15                    p                      8                      4

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 3 6 10 5 8 40 7 15 80 9 p 264 11 8 7p 13 4 80 Total $\sum f_{i}$ = 41 + p $\sum f_{i}x_{i}$ = 303 + 9p

We have:

$\sum f_{i}$ = 41 + p, $\sum f_{i}x_{i}$ = 303 + 9p

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

=> 7.68 = $\frac{303+9p}{41+p}$

=> 303 + 9p = 314.88 + 7.68p

=> 1.32p = 314.88 – 303

=> p = $\frac{11.88}{1.32}$

=> p = 9.

Q12) Find the value of p, if the mean of the following distribution is 20

x:                     15                    17                    19                    20+p                23

f:                      2                      3                      4                      5p                    6

Solution:

Calculation of Mean

 $x_{i}$ $f_{i}$ $x_{i}f_{i}$ 15 2 30 17 3 51 19 4 76 20 + p 5p (20 + p) 5p 23 6 138 Total $\sum f_{i}$ = 15 + 5p $\sum f_{i}x_{i}$ = 295 + (20+p) 5p

We have:

$\sum f_{i}$ = 15 + 5p, $\sum f_{i}x_{i}$ = 295 + (20+p) 5p

∴ Mean = $\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

=> 20 = $\frac{(295 + (20+p) 5p)}{15+5p}$

=> 295 + 100p + 5p2 = 300 + 100p

=> 5p2 = 300 – 295

=> 5p2 = 5

=> p2 = 1

=> p = 1.