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Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.3
  • Last Updated : 11 Feb, 2021

Question 1. Compute the mean deviation from the median of the following distribution:

Class0-1010-2020-3030-4040-50
Frequency51020510

Solution:

Calculating the median:

Median is the middle term of the observation in ascending order, Xi,

Here, the middle term is 25.

Class IntervalxifiCumulative Frequency|di| = |xi – M|fi |di|
0-1055520100
10-2015101510100
20-3025203500
30-40355911050
40-50451010120200
  Total = 50  Total = 450

Therefore, Median = 25



Mean Deviation, MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|

MD= 1/50 × 450

= 9

Therefore, mean deviation is 9.

Question 2. Find the mean deviation from the mean for the following data:

(i)

Classes0-100100-200200-300300-400400-500500-600600-700700-800
Frequencies489107543

(ii)

Classes95-105105-115115-125125-135135-145145-155
Frequencies91316263012

Solution:

(i) Mean = \frac{\sum f_ix_i}{f_i}



Mean = 17900/50

= 358

Also, the number of observations, N=50

Class IntervalxifiCumulative Frequency|di| = |xi – M|fi |di|
0-1005042003081232
100-200150812002081664
200-30025092250108972
300-400350103500880
400-5004507315092644
500-60055052750192960
600-700650426002921168
700-800750322503921176
  Total = 50Total = 17900 Total = 7896

MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|

= 1/50 × 7896

= 157.92

Therefore, mean deviation is 157.92.

(ii) Mean = \frac{\sum f_ix_i}{f_i}

Mean = 13630/106

= 128.58

Also, the number of observations, N=106

Class IntervalxifiCumulative Frequency|di| = |xi – M|fi |di|
95-105100990028.58257.22
105-11511013143018.58241.54
115-1251201619208.58137.28
125-1351302633801.4236.92
135-14514030420011.42342.6
145-15515012180021.42257.04
  N = 106Total = 13630 Total = 1272.6

MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|

= 1/106 × 1272.6

= 12.005

Question 3. Compute mean deviation from mean of the following distribution:

Marks10-2020-3030-4040-5050-6060-7070-8080-90
No. of students8101525201895

Solution:

Mean = \frac{\sum f_ix_i}{f_i}

= 5390/110

= 49

Class IntervalxifiCumulative Frequency|di| = |xi – M|fi |di|
10-2015812034272
20-30251025024240
30-40351552514210
40-50452511254100
50-60552011006120
60-706518117016288
70-8075967526234
80-9085542536180
  N = 110Total = 5390 Total = 1644

MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|

= 1/110 × 1644

= 14.94

Therefore, mean deviation is 14.94.

Question 4. The age distribution of 100 life-insurance policyholders is as follows:

Age (on nearest birthday)17-19.520-25.526-35.536-40.541-50.551-55.556-60.561-70.5
No. of persons5161226141265

Calculate the mean deviation from the median age.

Solution:

Number of observations, N = 96

So, N/2 = 96/2 = 48

The cumulative frequency just greater than 48 is 59, and therefore, the corresponding value of x is 38.25

Hence, Median = 38.25

Class IntervalxifiCumulative Frequency|di| = |xi – M|fi |di|
17-19.518.255520100
20-25.522.75162115.5248
36-35.530.7512337.590
36-40.538.25265900
41-50.545.7514737.5105
51-55.553.25128515180
56-60.558.2569120120
61-70.565.7559627.5137.5
  Total = 96  Total = 980.5

Number of observations, N = 96.

MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|

= 1/96 × 980.5

= 10.21

∴ The mean deviation is 10.21

Question 5. Find the mean deviation from the mean and from a median of the following distribution:

Marks0-1010-2020-3030-4040-50
No. of students5815166

Solution:

Number of observations, N = 50

So, N/2 = 50/2 = 25

The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28.

Therefore, Median = 28

Now,

Mean = \frac{\sum f_ix_i}{f_i}

= 1350/50

= 27

Class IntervalxifiCumulative Frequency|di| = |xi – Median|fi |di|FiXi|Xi – Mean|Fi |Xi – Mean|
0-10555231152522110
10-2015813131041201296
20-30251528345375230
30-4035164471125608128
40-50456501710227018108
  N = 50  Total = 478Total = 1350 Total = 472

Mean deviation from the median of observation = 478/50 = 9.56 

And, Mean deviation from mean of observation = 472/50 = 9.44 

∴ The mean deviation from the median is 9.56 and from the mean is 9.44.

Question 6. Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age16-2021-2526-3031-3536-4041-4545-5050-55
Number of persons5612142612169

Solution:

Converting the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

Age xifiCumulative Frequency|di| = |xi –38|fi |di|
15.5-20.5185520100
20.5-25.5236111590
25.5-30.528122310120
30.5-35.5331437570
35.5-40.538266300
40.5-45.5431275560
45.5-50.548169110160
50.5-55.553910015135
  N = 100  Total = 735

We have, N = 100 

So, N/2 = 100/2 = 50

The cumulative frequency just greater than N/2 is 63 and the corresponding class is 35.5-40.5.

l=35.5, f=26, h= 5, F =37

Therefore, Median = l + (N/2 – F)/f * h  = 35.5 + 50-37/26 * 5 =38

MD = \frac{1}{n}\sum f_i|d_i| = \frac{735}{100} = 73.5

Question 7.  Calculate the mean deviation from the median of the following data:

Class interval0-44-88-1212-1616-20
Frequency46852

Solution:

 fixifixi|xi-9.2|fi|xi-9.2|
0-44287.228.8
4-866363.219.2
8-12810800.86.4
12-16514704.824.0
16-20218368.817.6
 N=25 Total=230 total = 96.0

Mean = 230/25 =9.2

Mean Deviation=96/25 = 3.84

Question 8. Calculate the mean deviation from the median of the following data:

Class interval0-66-1212-1818-2424-30
Frequency45362

Solution:

 fixifixi|xi-14.1|fi|xi-14.1|
0-6431211.144.4
6-1259455.125.5
12-18315450.92.7
18-246211266.941.4
24-302275412.925.8
 N=20 Total=282 total = 139.8

Mean = 282/20 =14.1

Mean Deviation= 139.8/20 = 6.99

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