Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.3

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

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	10	20	5	10

Solution:

Calculating the median:

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

Here, the middle term is 25.

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
0-10	5	5	5	20	100
10-20	15	10	15	10	100
20-30	25	20	35	0	0
30-40	35	5	91	10	50
40-50	45	10	101	20	200
		Total = 50			Total = 450

Therefore, Median = 25

Mean Deviation, 

MD= 1/50 × 450

= 9

Therefore, mean deviation is 9.

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

(i)

Classes	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Frequencies	4	8	9	10	7	5	4	3

(ii)

Classes	95-105	105-115	115-125	125-135	135-145	145-155
Frequencies	9	13	16	26	30	12

Solution:

(i)

Mean = 17900/50

= 358

Also, the number of observations, N=50

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
0-100	50	4	200	308	1232
100-200	150	8	1200	208	1664
200-300	250	9	2250	108	972
300-400	350	10	3500	8	80
400-500	450	7	3150	92	644
500-600	550	5	2750	192	960
600-700	650	4	2600	292	1168
700-800	750	3	2250	392	1176
		Total = 50	Total = 17900		Total = 7896

= 1/50 × 7896

= 157.92

Therefore, mean deviation is 157.92.

(ii)

Mean = 13630/106

= 128.58

Also, the number of observations, N=106

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
95-105	100	9	900	28.58	257.22
105-115	110	13	1430	18.58	241.54
115-125	120	16	1920	8.58	137.28
125-135	130	26	3380	1.42	36.92
135-145	140	30	4200	11.42	342.6
145-155	150	12	1800	21.42	257.04
		N = 106	Total = 13630		Total = 1272.6

= 1/106 × 1272.6

= 12.005

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

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90
No. of students	8	10	15	25	20	18	9	5

Solution:

= 5390/110

= 49

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
10-20	15	8	120	34	272
20-30	25	10	250	24	240
30-40	35	15	525	14	210
40-50	45	25	1125	4	100
50-60	55	20	1100	6	120
60-70	65	18	1170	16	288
70-80	75	9	675	26	234
80-90	85	5	425	36	180
		N = 110	Total = 5390		Total = 1644

= 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)	17-19.5	20-25.5	26-35.5	36-40.5	41-50.5	51-55.5	56-60.5	61-70.5
No. of persons	5	16	12	26	14	12	6	5

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 Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
17-19.5	18.25	5	5	20	100
20-25.5	22.75	16	21	15.5	248
36-35.5	30.75	12	33	7.5	90
36-40.5	38.25	26	59	0	0
41-50.5	45.75	14	73	7.5	105
51-55.5	53.25	12	85	15	180
56-60.5	58.25	6	91	20	120
61-70.5	65.75	5	96	27.5	137.5
		Total = 96			Total = 980.5

Number of observations, N = 96.

= 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:

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	5	8	15	16	6

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,

= 1350/50

= 27

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – Median|	fi |di|	FiXi	|Xi – Mean|	Fi |Xi – Mean|
0-10	5	5	5	23	115	25	22	110
10-20	15	8	13	13	104	120	12	96
20-30	25	15	28	3	45	375	2	30
30-40	35	16	44	7	112	560	8	128
40-50	45	6	50	17	102	270	18	108
		N = 50			Total = 478	Total = 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:

Age	16-20	21-25	26-30	31-35	36-40	41-45	45-50	50-55
Number of persons	5	6	12	14	26	12	16	9

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	xi	fi	Cumulative Frequency	|di| = |xi –38|	fi |di|
15.5-20.5	18	5	5	20	100
20.5-25.5	23	6	11	15	90
25.5-30.5	28	12	23	10	120
30.5-35.5	33	14	37	5	70
35.5-40.5	38	26	63	0	0
40.5-45.5	43	12	75	5	60
45.5-50.5	48	16	91	10	160
50.5-55.5	53	9	100	15	135
		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

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

Class interval	0-4	4-8	8-12	12-16	16-20
Frequency	4	6	8	5	2

Solution:

	fi	xi	fixi	|xi-9.2|	fi|xi-9.2|
0-4	4	2	8	7.2	28.8
4-8	6	6	36	3.2	19.2
8-12	8	10	80	0.8	6.4
12-16	5	14	70	4.8	24.0
16-20	2	18	36	8.8	17.6
	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 interval	0-6	6-12	12-18	18-24	24-30
Frequency	4	5	3	6	2

Solution:

	fi	xi	fixi	|xi-14.1|	fi|xi-14.1|
0-6	4	3	12	11.1	44.4
6-12	5	9	45	5.1	25.5
12-18	3	15	45	0.9	2.7
18-24	6	21	126	6.9	41.4
24-30	2	27	54	12.9	25.8
	N=20		Total=282		total = 139.8

Mean = 282/20 =14.1

Mean Deviation= 139.8/20 = 6.99