Statistics – Exercise 15.1 – Class XI

1. Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17

Solution:

Given data is 4, 7, 8, 9, 10, 12, 13, 17

Mean of the data

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2. Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:

Given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data

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3.Find the mean deviation about the median for the data 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:

Given data is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

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4. Find the mean deviation about the median for the data

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:

Given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

The number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

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5. Find the mean deviation about the mean for the data

xi	5	10	15	20	25
fi	7	4	6	3	5

Solution:

xi	fi	fixi	|xi–x|	fi|xi–x|
5	7	35	9	63
10	4	40	4	16
15	6	90	1	6
20	3	60	6	18
25	5	125	11	55
	25	350		158

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6. Find the mean deviation about the mean for the data

xi	10	30	50	70	90
fi	4	24	28	16	8

Solution:

xi	fi	fi xi	|xi – x|	fi|xi – x|
10	4	40	40	160
30	24	720	20	480
50	28	1400	0	0
70	16	1120	20	320
90	8	720	40	320
	80	4000		1280

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7. Find the mean deviation about the mean for the data

xi	5	7	9	10	12	15
fi	8	6	2	2	2	6

Solution:

xi	fi	c.f
5	8	8
7	6	14
9	2	16
10	2	18
12	2	20
15	6	26

N = 26, which is even.

Median is the mean of 13th and 14th observations. Both of these observations lie in the  cumulative frequency 14, for which the corresponding observation is 7.

|xi – M|	2	0	2	3	5	8
fi	8	6	2	2	2	6
fi|xi – M|	16	0	4	6	10	48

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8. Find the absolute value of the derivations from median, for the data

xi	15	21	27	30	35
fi	3	5	6	7	8

Solution:

xi	fi	c.f
15	3	3
21	5	8
27	6	14
30	7	21
35	8	29

N = 29, which is an  odd number.

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9. Find the mean deviation about the mean for the data

Income per day	Number of persons
0-100	4
100-200	8
200-300	9
300-400	10
400-500	7
500-600	5
600-700	4
700-800	3

Solution:

Income per day	Number of persons	xi	fixi	|xi –x|	fi|xi –x|
0-100	4	50	200	308	1232
100-200	8	150	1200	208	1664
200-300	9	250	2250	108	972
300-400	10	350	3500	8	80
400-500	7	450	3150	92	644
500-600	5	550	2750	192	960
600-700	4	650	2600	292	1168
700-800	3	750	2250	392	1176
	50		17900		7896

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10. Find the mean deviation about the mean for the data

Height in cms	Number of boys
95-105	9
105-115	13
115-125	26
125-135	30
135-145	12
145-155	10

Solution:

Height in cms	Number of boys	xi	fixi	|xi – x|	fi| xi – x|
95-105	9	100	900	25.3	227.7
105-115	13	110	1430	15.3	198.9
115-125	26	120	3120	5.3	137.8
125-135	30	130	3900	4.7	141
135-145	12	140	1680	14.7	176.4
145-155	10	150	1500	24.7	247
			12530		1128.8

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11. Find the mean deviation about median for the following data:

Marks	Number of girls
0-10	6
10-20	8
20-30	14
30-40	16
40-50	4
50-60	2

Solution:

Marks	Number of girls	Cumulative frequency(c.f)	xi	|xi – median|	fi|xi –x|
0-10	6	6	5	22.85	137.1
10-20	8	14	15	12.85	102.8
20-30	14	28	25	2.85	39.9
30-40	16	44	35	7.15	114.4
40-50	4	48	45	17.15	68.6
50-60	2	50	55	27.15	54.3
	50				517.1

The class interval containing the (^N/₂) or 25^th item is 20-30

Thus, 20-30 is the median class

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

age	Number
16-20	5
21-25	6
26-30	12
31-35	14
36-40	26
41-45	12
46-50	16
51-55	9

Solution:

We need to convert the given dat into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each interval in order to make it continuous.

age	Number	c.f	xi	|xi – x|	fi|xi–x|
16-20	5	5	18	20	100
21-25	6	11	23	15	90
26-30	12	23	28	10	120
31-35	14	37	33	5	70
36-40	26	63	38	0	0
41-45	12	75	43	5	60
46-50	16	91	48	10	160
51-55	9	100	53	15	135
	100				735

The class interval containing the ^N/₂ ^th or 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

We know that,

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