RD Sharma solution class 7 chapter 23 Data Handling II(Central values) Exercise 23.2

Exercise 23.2

Page-23.12

Question 1:

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.

Answer 1:

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 (xi)	Frequency (fi)	xifi
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 $\sum f_i{x}_i =73$
∴ The mean score = $\frac{{\sum }_{}{f}_i{x}_i}{\sum f_i} = \frac{73}{20} = 3.65$.

Question 2:

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.

Answer 2:

The frequency table for the given data is as follows:
          Wages ( x_i ) :                               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

xi	fi	fi xi
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 $= \frac{\sum _{}^{}{f}_i x_i}{\sum _{}^{}{f}_i } = \frac{2540}{15} = \mathrm{Rs}. 169.33$.

Question 3:

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.

Answer 3:

                                    Calculation of Mean

xi	fi	fi xi
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 = ​$\frac{\sum _{}^{}{f}_i{x}_i}{\sum _{}^{}{f}_i} = \frac{975}{15} = 65 \mathrm{kg}$.

Question 4:

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.

Answer 4:

        Calculation of Mean

xi	fi	fi xi
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 age = ​$\frac{\sum _{}^{}{f}_i{x}_i}{\sum _{}^{}{f}_i} = \frac{770}{50} = 15.4 \mathrm{yrs}$.

Question 5:

Calculate the mean for the following distribution:

x :	5	6	7	8	9
f :	4	8	14	11	3

Answer 5:

                       Calculation of Mean

xi	fi	fi xi
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 = ​$\frac{\sum _{}^{}{f}_i{x}_i}{\sum _{}^{}{f}_i} = \frac{281}{40} = 7.025$.

Page-23.13

Question 6:

Find the mean of the following data:

x:	19	21	23	25	27	29	31
f:	13	15	16	18	16	15	13

Answer 6:

                    Calculation of Mean

xi	fi	fixi
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  $=\frac{\sum _{}{f}_i{x}_i}{\sum _{}{f}_i} = \frac{2650}{106} = 25$.

Question 7:

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

Answer 7:

            Calculation of Mean

xi	fi	fi xi
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: 

${\sum }_{}{f}_i$ = 50, ​${\sum }_{}{f}_i{x}_i$ = 530 +25p

∴ Mean =$\frac{{\sum }_{}{f}_i{x}_i}{{\sum }_{}{f}_i}$  

$$\begin{gathered} \Rightarrow 20.6 = \frac{530 +25p}{50} \\ \Rightarrow 20.6\times 50 = 530 +25p\Rightarrow 1030 = 530 +25p\Rightarrow 1030 - 530 = 25p\Rightarrow 500 = 25p\Rightarrow p = \frac{500}{25}\Rightarrow p = 20 \end{gathered}$$ 

Question 8:

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

x:	5	10	15	20	25
f:	6	p	6	10	5

Answer 8:

                                 Calculation of Mean

xi	fi	fi xi
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 =$\frac{{\sum }_{}{f}_i{x}_i}{{\sum }_{}{f}_i}$  

$$\begin{gathered} \Rightarrow 15= \frac{445 +10p}{27 +p} \\ \Rightarrow 15 (27 +p) = 445 +10p\Rightarrow 405 + 15p =445 +10p\Rightarrow 15p - 10p = 445 -405\Rightarrow 5p = 40\Rightarrow p = 40÷5 \end{gathered}$$

Therefore, p =  8.

Question 9:

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

Answer 9:

​                      Calculation of Mean

xi	fi	fixi
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  $=\frac{\sum _{}{f}_i{x}_i}{\sum _{}{f}_i}$

$$\begin{gathered} \Rightarrow 16.6 = \frac{1228 +24p}{100} \\ \Rightarrow 16.6\times 100 = 1228 +24p \\ \Rightarrow 1660 = 1228 +24p \\ \Rightarrow 1660 - 1228 = 24p \\ \Rightarrow 432 = 24p \\ \Rightarrow p = \frac{432}{24} \\ \Rightarrow p =18. \end{gathered}$$

Question 10:

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

Answer 10:

​                      Calculation of Mean

xi	fi	fixi
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  $=\frac{\sum _{}{f}_i{x}_i}{\sum _{}{f}_i}$

$$\begin{gathered} \Rightarrow 12.58 = \frac{524 +7p}{50} \\ \Rightarrow 12.58\times 50 = 524 +7p \\ \Rightarrow 629 = 524 +7p \\ \Rightarrow 629 - 524 = 7p \\ \Rightarrow 105 = 7p \\ \Rightarrow p = \frac{105}{7} \\ \Rightarrow p =15. \end{gathered}$$

Question 11:

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

Answer 11:

                                 Calculation of Mean

xi	fi	fi xi
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
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 =$\frac{{\sum }_{}{f}_i{x}_i}{{\sum }_{}{f}_i}$  

$$\begin{gathered} 7.68 = \frac{303 +9p}{41+p} \\ \Rightarrow 7.68 \times (41 +p) =303 +9p\Rightarrow 314.88 + 7.68p = 303 +9p \\ \Rightarrow 314.88 -303 = 9p -7.68p \\ \Rightarrow 11.88 = 1.32p \\ \Rightarrow p = \frac{11.88}{1.32} \\ \Rightarrow p =9. \end{gathered}$$

Question 12:

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	5 p	6

Answer 12:

                                 Calculation of Mean

xi	fi	fi xi
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 =$\frac{{\sum }_{}{f}_i{x}_i}{{\sum }_{}{f}_i}$  

$$\begin{gathered} 20= \frac{295 + (20+p)5p}{15+5p} \\ \Rightarrow 20 \times (15 +5p) =295 + (20+p)5p\Rightarrow 300+ 100p = 295 +100p + 5p \end{gathered}$$

⇒ 300 - 295 + 100p -100p = 5p²
⇒ 5 = 5p²
⇒ p² = 1
⇒ p = 1.