Statistics - Solutions

 CBSE Class–11 Mathematics

NCERT Solutions
Chapter - 15 Statistics
Exercise 15.1

---

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

Ans. Mean of the given data 

Mean deviation about mean =  =  = 3

---

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Ans. Mean of the given data 

Mean deviation about mean = 

=  = 8.4

---

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Ans. Arranging the data in ascending order, 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Here,  (even number)

Median = 

=  = 13.5

Mean deviation about median = 

=  = 2.33

---

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Ans. Arranging the data in ascending order, 36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Here,  (even number)

Median=  =  = 47.5

Mean deviation about median = 

=  = 7

---

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5.

Ans.

Mean 

=  = 14

Mean deviation about mean = 

= 15825$\frac{158}{25}$ = 6.32

---

6.

Ans.

Mean 

=  = 50

Mean deviation about mean = 

=  = 16

---

Find the mean deviation about the median for the data in Exercises 7 and 8.

7.

Ans.

Here,N = 26 (even number)

Therefore, Median ==262thterm+(262+1)thterm2=13thterm+14thterm2=7+72=7$=\frac{{\frac{26}{2}}^{th}term+{\left(\frac{26}{2}+1\right)}^{th}term}{2}=\frac{{13}^{th}term+{14}^{th}term}{2}=\frac{7+7}{2}=7$ 

Median = 7

Mean deviation about median = 

=  = 3.23

---

8.

Ans.

Here,N = 29 (odd number) 

Median =(29+12)thterm=15thterm=30${\left(\frac{29+1}{2}\right)}^{th}term={15}^{th}term=30$

Median = 30

Mean deviation about median = 

=  = 5.1

---

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.

Ans.

Mean 

=  = 358

Mean deviation about mean = 

=  = 157.92

---

10.

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

Ans.

Mean 

=  = 125.3

Mean deviation about mean = 

=  = 11.288

---

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

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

Ans.

Marks	Mid Values (xi)	fi	cf	|xi−M|$\left|x_i^{}-M\right|$	fi|xi−M|$f_i^{}\left|x_i^{}-M\right|$
0-10	5	6	6	22.86	137.16
10-20	15	8	14	12.86	102.88
20-30	25	14	28	2.86	40.04
30-40	35	16	44	7.14	114.24
40-50	45	4	48	17.14	68.56
50-60	55	2	50	27.14	54.28
Total		50			517.16

Here,  = 25

Median class = 20 – 30

Median (M) = 20+25−1414×10$20+\frac{25-14}{14}\times 10$

M = 20 + 7.86 = 27.86

Mean deviation about median = 

=  = 10.34

---

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

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

Ans. Firstly the data is to be made continous by making classes 15.5-20.5, 20.5-25.5, 25.5-30.5, 30.5-35.5, 35.5-40.5, 40.5-45.5, 45.5-50.5, 50.5-55.5

otal

Age	Mid Value (xi)	fi	cf	|xi−M|$\left|x_i^{}-M\right|$	fi|xi−M|$f_i^{}\left|x_i^{}-M\right|$
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
Total		100			735

Here,  = 50

Median class = 35.5 – 40.5

Median = 

= 35.5 + 2.5 = 38

Mean deviation about median = 

=  = 7.35