RD Sharma Class 10 Solutions Chapter 7 Statistics

RD Sharma Class 10 Solutions Chapter 7 Statistics

RD Sharma Class 10 Solutions Statistics Exercise 7.1

Question 1.
Calculate the mean for the following distribution :

X	5	6	7	8	9
f	4	8	14	11	3

Solution:

Question 2.
Find the mean of the following data:

X	19	21	23	25	27	29	31
f	13	15	16	18	16	15	13

Solution:

Question 3.
If the mean of the following data is 20.6. Find the value of p. (C.B.S.E. 1997)

X	10	15	p	25	35
y	3	10	25	7	5

Solution:

Question 4.
If the mean of the following data is 15, find p. (C.B.S.E. 1992C)

X	5	10	15	20	25
f	6	P	6	10	5

Solution:

Question 5.
Find the value of p for the following distribution whose mean is 16.6.

X	8	12	15	P	20	25	30
f	12	16	20	24	16	8	4

Solution:
Mean = 16.6

Question 6.
Find the missing value of p for the following distribution whose mean is 12.58. (C.B.S.E. 1992C)

X	5	8	10	12	P	20	25
f	2	5	8	22	7	4	2

Solution:

Question 7.
Find the missing frequency (p) for the following distribution whose mean is 7.68.

X	3	5	7	9	11	13
f	6	8	15	P	8	4

Solution:

Question 8.
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

Age (in years)	15	16	17	18	19	20
No. of students	3	8	10	10	5	4

Solution:

Question 9.
Candidates of four schools appear in a mathematics test. The data were as follows :

Schools	No. of Candidates	Average Score
I	60	75
II	48	80
III	Not available	55
IV	40	50

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Solution:

Question 10.
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss	No. of tosses
0	38
1	144
2	342
3	287
4	164
5	25
Total	1000

Solution:

Question 11.
The arithmetic mean of the following data is 14, find the value of k. (C.B.S.E. 2002C)

X	5	10	15	20	25
f	7	k	8	4	5

Solution:
Mean=14

⇒ 14 (24 + k) = 360 + 10k
⇒ 336 + 14k = 360 + 10k
⇒ 14k- 10k- 360 -336 24
⇒ 4k = 24
⇒ k= \(\frac { 24 }{ 4 }\) = 6 4
Hence k = 6

Question 12.
The arithmetic mean of the following data is 25, find the value of k. (C.B.S.E. 2001)

X	5	15	25	35	45
f	3	k	3	6	2

Solution:
Mean =25

⇒ 25 (14 + k) = 390 + 15k
⇒ 350 + 25k= 390 + 15k
⇒ 25k- 15k = 390 -350
⇒ 10k = 40 ⇒ k = \(\frac { 40 }{ 10 }\) = 4
Hence k = 4

Question 13.
If the mean of the following data is 18.75. Find the value of p.

X	10	15	P	25	30
f	5	10	7	8	2

Solution:

⇒ 460 + 7p = 32 (18.75)
⇒ 460 + 7p = 600
⇒ 7p = 600 – 460 = 140
⇒ p =  \(\frac { 140 }{ 7 }\) = 20
∴ p = 20

Question 14.
Find the value of p, if the mean of the following distribution is 20.

X	15	17	19	20 + p	23
f	2	3	4	5p	6

Solution:

⇒ 5p² + 100p + 295 = 20 (15 + 5p)
⇒ 5p² + 100p + 295 = 300 + 100p
⇒ 5p² + 100p – 100p = 300 – 295
⇒  5p² = 5 ⇒  p²  =  \(\frac { 5 }{ 5 }\)  = 1
⇒ P= ±1
P = -1 i s not possible
∴ p= 1

Question 15.
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

Solution:

RD Sharma Class 10 Solutions Chapter 7 Statistics Ex 7.1

RD Sharma Class 10 Solutions

• Chapter 7 Statistics Ex 7.1
• Chapter 7 Statistics Ex 7.2
• Chapter 7 Statistics Ex 7.3
• Chapter 7 Statistics Ex 7.4
• Chapter 7 Statistics Ex 7.5
• Chapter 7 Statistics Ex 7.6