RS AGGARWAL CLASS 9 CHAPTER 18 MEAN, MEDIAN AND MODE OF UNGROUPED DATA EXERCISE 18B

 EXERCISE 18B

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Question 1:

Obtain the mean of the following distribution:

Variable (xi)	4	6	8	10	12
Frequency (fi)	4	8	14	11	3

Answer 1:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Variable (xi)	4	6	8	10	12
Frequency (fi)	4	8	14	11	3

Mean = $\frac{\left(4\times 4\right)+\left(6\times 8\right)+\left(8\times 14\right)+\left(10\times 11\right)+\left(12\times 3\right)}{4+8+14+11+3}$
          = $\frac{16+48+112+110+36}{40}$
          = $\frac{322}{40}$
          = 8.05

Hence, the mean of the following distribution is 8.05 .

Question 2:

The following table shows the weights of 12 workers in a factory:

Weight (in kg)	60	63	66	69	72
No. of workers	4	3	2	2	1

Find the mean weight of the workers.

Answer 2:

We will make the following table:
 

Weight (xi)	No. of Workers (fi)	(fi)(xi)
60	4	240
63	3	189
66	2	132
69	2	138
72	1	72
	$\sum _{}^{}{f}_i =$12	$\sum _{}^{}{f}_i x_i =$771

Thus, we have:

$\mathrm{Mean} = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$
= $\frac{771}{12}= 64.25 \mathrm{kg}$

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Question 3:

The measurements (in mm) of the diameters of the heads of 50 screws are given below:

Diameter (in mm) (xi)	34	37	40	43	46
Number of screws (fi)	5	10	17	12	6

Calculate the mean diameter of the heads of the screws.

Answer 3:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Diameter (in mm) (xi)	34	37	40	43	46
Number of screws (fi)	5	10	17	12	6

Mean = $\frac{\left(34\times 5\right)+\left(37\times 10\right)+\left(40\times 17\right)+\left(43\times 12\right)+\left(46\times 6\right)}{5+10+17+12+6}$
          = $\frac{170+370+680+516+276}{50}$
          = $\frac{2012}{50}$
          = 40.24

Hence, the mean diameter of the heads of the screws is 40.24 .

Question 4:

The following data give the number of boys of a particular age in a class of 40 students.

Age (in years)	15	16	17	18	19	20
Frequency (fi)	3	8	9	11	6	3

Calculate the mean age of the students.

Answer 4:

We will make the following table:
 

Age (xi)	Frequency (fi)	(fi)(xi)
15	3	45
16	8	128
17	9	153
18	11	198
19	6	114
20	3	60
	$\sum _{}^{}{f}_i =$40	$\sum f_i{x}_i=$698

Thus, we have:
$\mathrm{Mean} = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$
$= \frac{698}{40}= 17.45 \mathrm{years}$

Question 5:

Find the mean of the following frequency distribution:

Variable (xi)	10	30	50	70	89
Frequency (fi)	7	8	10	15	10

Answer 5:

We will make the following table:
 

Variable (xi)	Frequency (fi)	(fi)(xi)
10	7	70
30	8	240
50	10	500
70	15	1050
89	10	890
	$\sum _{}^{}{f}_i =$50	$\sum _{}^{}{f}_i x_i =$2750

Thus, we have:

$\mathrm{Mean} = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$
$= \frac{2750}{50} = 55$

Question 6:

Find the mean of daily wages of 40 workers in a factory as per data given below:

Daily wages (in ₹) (xi)	250	300	350	400	450
Number of workers (fi)	8	11	6	10	5

Answer 6:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Daily wages (in ₹) (xi)	250	300	350	400	450
Number of workers (fi)	8	11	6	10	5

Mean = $\frac{\left(250\times 8\right)+\left(300\times 11\right)+\left(350\times 6\right)+\left(400\times 10\right)+\left(450\times 5\right)}{8+11+6+10+5}$
          = $\frac{2000+3300+2100+4000+2250}{40}$
          = $\frac{13650}{40}$
          = 341.25

Hence, the mean of daily wages of 40 workers in a factory is 341.25 .

Question 7:

If the mean of the following data is 20.2, find the value of  p.

Variable (xi)	10	15	20	25	30
Frequency (fi)	6	8	p	10	6

Answer 7:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Variable (xi)	10	15	20	25	30
Frequency (fi)	6	8	p	10	6

Mean = $\frac{\left(10\times 6\right)+\left(15\times 8\right)+\left(20\times p\right)+\left(25\times 10\right)+\left(30\times 6\right)}{6+8+p+10+6}$
$$\begin{gathered} \Rightarrow 20.2=\frac{60+120+20p+250+180}{30+p} \\ \Rightarrow 20.2\left(30+p\right)=610+20p \\ \Rightarrow 606+20.2p=610+20p \\ \Rightarrow 20.2p-20p=610-606 \\ \Rightarrow 0.2p=4 \\ \Rightarrow p=\frac{4}{0.2} \\ \Rightarrow p=\frac{40}{2} \\ \Rightarrow p=20 \end{gathered}$$

Hence, the value of  p is 20.

Question 8:

If the mean of the following frequency distribution is 8, find the value of p.

x	3	5	7	9	11	13
f	6	8	15	p	8	4

Answer 8:

We will make the following table:
 

(xi)	(fi)	(fi)(xi)
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
	$\sum _{}^{}{f}_i =$41 + p	$\sum _{}^{}{f}_i x_i =$303 + 9p

We know:

$\mathrm{Mean} = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$
Given:
Mean = 8
Thus, we have:

$$\begin{gathered} 8 = \frac{303+9p}{41+p} \\ \Rightarrow 328+8p = 303+9p \\ \Rightarrow p = 25 \end{gathered}$$

Question 9:

Find the missing frequency p for the following frequency distribution whose mean is 28.25.

x	15	20	25	30	35	40
f	8	7	p	14	15	6

Answer 9:

We will prepare the following table:
 

(xi)	(fi)	(fi)(xi)
15	8	120
20	7	140
25	p	25p
30	14	420
35	15	525
40	6	240
	$\sum _{}^{}{f}_i =$50 + p	$\sum _{}^{}{f}_i x_i =$1445 + 25p

Thus, we have:
$\mathrm{Mean} = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$
 $$\begin{gathered} \Rightarrow 28.25 = \frac{1445+25p}{50+p} \\ \Rightarrow 28.25\left(50+p\right) = \left(1445+25p\right) \\ \Rightarrow 1412.5 + 28.25p = 1445+25p \\ \Rightarrow 3.25 p = 32.5 \\ \Rightarrow p =10 \end{gathered}$$

Question 10:

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

x	8	12	15	p	20	25	30
f	12	16	20	24	16	8	4

Answer 10:

We will make the following table:
 

(xi)	(fi)	(fi)(xi)
8	12	96
12	16	192
15	20	300
p	24	24p
20	16	320
25	8	200
30	4	120
	$\sum _{}^{}{f}_i =$100	$\sum _{}^{}{f}_i x_i =$1228 + 24p

Thus, we have:

$\mathrm{Mean} = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$
$$\begin{gathered} \Rightarrow 16.6 = \frac{\left(1228+24p\right)}{100} \\ \Rightarrow 16.6\times 100 = \left(1228+24p\right) \\ \Rightarrow 1660 = 1228+24p \end{gathered}$$
$$\begin{gathered} \Rightarrow 24p=432 \\ \Rightarrow p=18 \end{gathered}$$

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Question 11:

Find the missing frequencies in the following frequency distribution whose mean is 34.

x	10	20	30	40	50	60	Total
f	4	f1	8	f2	3	4	35

Answer 11:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

x	10	20	30	40	50	60	Total
f	4	f1	8	f2	3	4	35

Mean = $\frac{\left(10\times 4\right)+\left(20\times f_1\right)+\left(30\times 8\right)+\left(40\times f_2\right)+\left(50\times 3\right)+\left(60\times 4\right)}{35}$
$$\begin{gathered} \Rightarrow 34=\frac{40+20f_1+240+40f_2+150+240}{35} \\ \Rightarrow 34\left(35\right)=670+20f_1+40f_2 \\ \Rightarrow 1190-670=20f_1+40f_2 \\ \Rightarrow 20f_1+40f_2=520 \\ \Rightarrow 20\left(f_1+2f_2\right)=520 \\ \Rightarrow f_1+2f_2=\frac{520}{20} \\ \Rightarrow f_1+2f_2=26 \\ \Rightarrow f_1=26-2f_2 ...\left(1\right) \end{gathered}$$

Also, 4 + f₁ + 8 + f₂ + 3 + 4 = 35
⇒ 19 + f₁ + f₂ = 35
⇒ f₁ + f₂ = 35 − 19
⇒ f₁ + f₂ = 16
⇒ 26 − 2f₂ + f ₂ = 16             (from (1))
⇒ 26 − f₂ = 16
⇒ 26 − 16 =  f₂
⇒  f₂ = 10

Putting the value of f₂ in (1), we get
f₁ = 26 − 2(10) = 6

Hence, the value of f₁ and  f₂ is 6 and 10, respectively.

Question 12:

Find the missing frequencies in the following frequency distribution, whose mean is 50.

x	10	30	50	70	90	Total
f	17	f1	32	f2	19	120

Answer 12:

We will prepare the following table:
 

(xi)	(fi)	(fi)(xi)
10	17	170
30	f1	30f1
50	32	1600
70	f2	70f2
90	19	1710
	$\sum _{}^{}{f}_i =$120	$\sum _{}^{}{f}_i x_i =$3480 + 30f1 + 70f2

Thus, we have:
$Mean = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$

$\Rightarrow 50 = \frac{3480+30f_1+70f_2}{120}$

_{$$\begin{gathered} \Rightarrow 6000=3480+30f_1+70f_2 \\ \Rightarrow 30f_1+70f_2 = 2520 ..... \left(\mathrm{i}\right) \end{gathered}$$}


Also,
Given:
17 + f₁ + 32 + f₂ + 19 = 120
⇒ 68 + f₁ + f₂ = 120
⇒ f₁ + f₂ = 52
or, f₂ = 52 $-$ f₁  ...(ii)
By putting the value of f₂ in (i), we get:
2520 = 30f₁ + 70(52 $-$ f₁)
⇒ 2520 = 30f₁ + 3640 $-$ 70f₁
⇒ 40f₁ = 1120
⇒ f₁ = 28
Substituting the value in (ii), we get:
f₂ = 52 $-$ f₁ = 52 $-$ 28 = 24

Question 13:

Find the value of p, when the mean of the following distribution is 20.

x	15	17	19	20 + p	23
f	2	3	4	5p	6

Answer 13:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

x	15	17	19	20 + p	23
f	2	3	4	5p	6

Mean = $\frac{\left(15\times 2\right)+\left(17\times 3\right)+\left(19\times 4\right)+\left(\left(20+p\right)\times 5p\right)+\left(23\times 6\right)}{2+3+4+5p+6}$
$$\begin{gathered} \Rightarrow 20=\frac{30+51+76+100p+5p^2+138}{15+5p} \\ \Rightarrow 20\left(15+5p\right)=5p^2+100p+295 \\ \Rightarrow 300+100p=5p^2+100p+295 \\ \Rightarrow 5p^2=300-295 \\ \Rightarrow 5p^2=5 \\ \Rightarrow p^2=\frac{5}{5} \\ \Rightarrow p^2=1 \\ \Rightarrow p=\pm 1 \end{gathered}$$

Hence, the value of  p is ±1.

Question 14:

The mean of the following distribution is 50.

x	10	30	50	70	90
f	17	5a + 3	32	7a – 11	19

Find the value of a and hence the frequencies of 30 and 70.

Answer 14:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

x	10	30	50	70	90
f	17	5a + 3	32	7a – 11	19

Mean = $\frac{\left(10\times 17\right)+\left(30\times \left(5a+3\right)\right)+\left(50\times 32\right)+\left(70\times \left(7a-11\right)\right)+\left(90\times 19\right)}{17+5a+3+32+7a-11+19}$
$$\begin{gathered} \Rightarrow 50=\frac{170+150a+90+1600+490a-770+1710}{60+12a} \\ \Rightarrow 50\left(60+12a\right)=2800+640a \\ \Rightarrow 3000+600a=2800+640a \\ \Rightarrow 640a-600a=3000-2800 \\ \Rightarrow 40a=200 \\ \Rightarrow a=\frac{200}{40} \\ \Rightarrow a=5 \end{gathered}$$

Hence, the value of a is 5.

Also, the frequency of 30 is 28 and the frequency of 70 is 24.