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Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.2

  • Last Updated : 11 Feb, 2021
Geek Week

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

Heights in inches585960616263646566
No. Of Students15203235352220108

Solution:

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

So, Median = 61

Let us assume, 



xi =Heights in inches

fi = Number of students

xifiCumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|
581515345
592035240
603267132
613510200
6235137135
6322159244
6420179360
6510189440
668197540
 N = 197  Total = 336

N=197

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/197 × 336

= 1.70



Therefore, the mean deviation is 1.70.

Question 2. The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:

Number of calls01234567
Frequency1421254351403912

Compute the mean deviation about the median.

Solution:

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

We know, Median is the even term, (3+5)/2 = 4

So, Median = 8

Let us assume

xi =Number of calls

fi = Frequency

N = 245

xifiCumulative Frequency

|di| = |xi – M|



= |xi – 61|

fi |di|
01414456
12135363
22560250
343103143
45115400
540194140
639233278
712245336
 Total = 245  Total = 366

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/245 × 336

= 1.49

Therefore, mean deviation is 1.49.

Question 3. Calculate the mean deviation about the median of the following frequency distribution:

xi57911131517
fi246810128

Solution:

Calculating the median,

We know, Number of observations, N = 50

Median = (50)/2 = 25

Therefore, the median corresponding to 25 is 13.



xifiCumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|
522816
746624
9612424
11820216
13103000
151242224
17850432
 Total = 50  Total = 136

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/50 × 136

= 2.72

Therefore, the mean deviation is 2.72.

Question 4. Find the mean deviation from the mean for the following data:

(i)

xi579101215
fi862226

(ii)

xi510152025
fi74635

(iii)

xi1030507090
fi42428168

Solution:



(i) We know,

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

xifiCumulative Frequency (xifi)|di| = |xi – Mean|fi |di|
5840432
7642212
921800
1022012
1222436
15690636
 Total = 26Total = 234 Total = 88

Now, Mean = 234/26

= 9

Mean Deviation = \frac{\sum f_i|d_i|}{f_i}

= 88/26

= 3.3

∴ The mean deviation is 3.3

(ii) We know,

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



xifiCumulative Frequency (xifi)|di| = |xi – Mean|fi |di|
5735963
10440416
1569016
20360618
2551251155
 Total = 25Total = 350 Total = 158

Mean = 350/25

= 14
Mean Deviation = \frac{\sum f_i|d_i|}{f_i}

= 158/25

= 6.32

∴ The mean deviation is 6.32

(iii) We know,

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

= 4000/80

= 50

xifiCumulative Frequency (xifi)|di| = |xi – Mean|fi |di|
1044040160
302472020480
5028140000
7016112020320
90872040320
 Total = 80Total = 4000 Total = 1280

Mean Deviation = \frac{\sum f_i|d_i|}{f_i}



= 1280/80

= 16

∴ The mean deviation is 16

Question 5. Find the mean deviation from the median for the following data :

(i)

xi15212730
fi3567

(ii)

xi74894254919435
fi201224534

(iii)

Marks obtained1011121415
No. of students23834

Solution:

(i) We know, 

Number of observations, N = 21

Median = (21)/2 = 10.5

Therefore, the median corresponding to 10.5 is 27

xifiCumulative Frequency|di| = |xi – M|fi |di|
15331545
2158945
27614318
3072100
 Total = 21Total = 46 Total = 108

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/21 × 108

= 5.14

∴ The mean deviation is 5.14

(ii) We know, 

Number of observations, N =50

Median = (50)/2 = 25

Therefore, the median corresponding to 25 is 74.

xifiCumulative Frequency|di| = |xi – M|fi |di|
7420439156
891263264
422102080
5443000
9154215180
943471785
354502060
 Total = 50Total = 189 Total = 625

MD=\frac{1}{n} \sum^{n}_{i=1}|di|



= 1/50 × 625

= 12.5

Therefore, the mean deviation is 12.5

(iii) We know, 

Number of observations, N =20

Median = (20)/2 = 10

So, the median corresponding to 10 is 12.

xifiCumulative Frequency|di| = |xi – M|fi |di|
102224
113513
1281300
1431626
15420312
 Total = 20  Total = 25

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/20 × 25

= 1.25

∴ The mean deviation is 1.25

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