NCERT Solutions For Class 11 Maths Chapter 15 Statistics Exercise 15.1

Class 11 Maths NCERT Solutions for Chapter 15 Statistics Exercise 15.1

Statistics Exercise 15.1 Solutions

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

Solution

The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, 

The deviations of the respective observations from the meanare –6, –3, –2, –1, 0, 2, 3, 7
The absolute values of the deviations, i.e.,are 6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is

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

Solution

The given data is
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data,

The deviations of the respective observations from the mean  are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6 
The absolute values of the deviations, i.e., , are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.
The required mean deviation about the mean is 

= 84/10 
= 8.4

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

Solution

The 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

The deviations of the respective observations from the median, i.e. x_i – M ,are
–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The absolute values of the deviations, |x_i – M |, are
3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The required mean deviation about the median is

= 28/12 = 2.33

4. Find the mean deviation about the median for the data
36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution

The given data is
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Here, 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

The deviations of the respective observations from the median, i.e. x_i – M are
–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5
The absolute values of the deviations,  |x_i – M |, are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Thus, the required mean deviation about the median is 

= 70/10 = 7

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
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

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	fixi
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

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

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

Solution

The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

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

Here, N = 26, which is even.
Median is the mean of 13^th and 14^th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e, |x_i – M|, are  

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

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

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

Solution

The given observations are already in ascending order. 

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table. 

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

Here, N = 29, which is odd. 
∴ Median = [(29 + 1)/2 ]^th observation = 15^th observation 
This observation lies in the cumulative frequency 21, for which the corresponding observation is  30. 
Median = 30 
The absolute values of the deviations from median, i.e. |x_i – M|, are  

|xi – M|	15	9	3	0	5
fi	3	5	6	7	8
fi |xi – M|	45	45	18	0	40

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

The following table is formed. 

Income per day	Number of person fi	Mid – point xi	fi xi
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

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

The following table is formed. 

Height in cms	Number of boys fi	Mid – point xi	fi xi
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

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

The following table is formed. 

Marks	Number of boys  fi	Cumulative frequency (c. f.)	Mid – point xi	|xi – Med.|	fi |xi – Med.|
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)^th or 25^th item is  20 - 30 .
Therefore, 20 - 30  is the median class. 
It is known that,  

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

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval. 

The table is formed as follows. 

Age	Number fi	Cumulative frequency (c.f.)	Mid – point xi	|xi – Med.|	fi |xi – Med.|
15.5 – 20.5	5	5	18	20	100
20.5 – 25.5	6	11	23	15	90
25.5 – 30.5	12	23	28	10	120
30.5 – 35.5	14	37	33	5	70
35.5 – 40.5	26	63	37	0	0
40.5 – 45.5	12	75	43	5	60
45.5 – 50.5	16	91	48	10	160
50.5 – 55.5	9	100	53	15	135
	100				735

The class interval containing the (N/2)^th or  50^th item is  35.5 - 40.5 . 
Therefore, 35.5 - 40.5 is the median class.
It is known that, 

Here, l = 35.5 , C = 37 , F = 26, h = 5, and N = 100