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

RD Sharma Class 7 Solutions Chapter 23 Ex 23.2 PDF Free Download

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.

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