RD Sharma Solutions Class 10 Statistics Exercise 7.1

RD Sharma Solutions Class 10 Chapter 7 Exercise 7.1

RD Sharma Class 10 Solutions Chapter 7 Ex 7.1 PDF Download

Exercise 7.1

 

Q.1: Calculate the mean for the following distribution:

 

x: 5              6              7              8              9
f: 4              8              14           11           3

 

Sol:

 

X f fx
5 4 20
6 8 48
7 14 98
8 11 88
9 3 27
  N = 40 281

Mean = 281/4 =7.025

 

2. Find the mean of the following data:

 

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


Soln:

X f fx
18 13 247
21 15 315
23 16 368
25 18 450
27 16 432
29 15 435
31 13 403
  N = 106 Sum = 2620

Mean (x) = 2680/106 = 25

 

3. If the mean of the following data is 20.6. Find the value of p.

 

x: 10           15           p             25           35
f: 3              10           25           7              5

 

Soln:

 

X f fx
10 3 30
5 10 150
P 25 25p
25 7 175
35 5 175
N = 106 Sum = 2620

Given

Mean = 20.6

(530+25p)/50 = 20.6

25p = 20.6

P = 20

 

4. If the mean of the following data is 15, find p

 

x: 5              10           15           20           25
f: 6              p             6              10           5

 

Soln:

X f fx
5 6 30
10 P 10p
15 6 90
20 10 200
25 5 125
N = p127 Sum = 10p + 445

Given

Mean =15

(10p + 445)/(p+27) = 15

10p + 445 = 15p +405

15p – 10p = 445 – 405

5p = 40

P = 8

 

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

 

Soln:

 

X f fx
8 12 96
12 12 192
15 20 300
P 24 24p
20 16 320
25 8 200
30 4 120
N = 100 Sum = 24p + 1228

 

Given

Mean = 16.6

(24p+1228)/100 = 16.6

24p + 1228 = 1660

24p = 1660 – 1228

P = 432/24

P = 18

 

6. Find the missing value of p for the following distribution whose mean is 12.58

 

x: 5              8          10            12               p             20           25

 

f: 2              5           8             22               7              4              2

 

 

Soln:

 

x f fx
5 2 10
8 5 40
10 8 80
12 22 264
P 7 7p
20 24 480
25 2 50
N = 50 Sum = 524 + 7p

 

Given mean = 12.58

Sum/N = 12.58

(524 +7p)/50 = 12.58

524 + 7p = 629

7p = 105

P = 15

 

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

 

Soln:

 

X f fx
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
N = P + 41 Sum 9p = 303

 

Given

Mean = 7.68

(7p+303)/p+41 =7.68

9p + 303 = p(7.68) + 314.88

9p – p(7.68) = 314.88 – 303

1.32p = 11.88

P = (11.88)/1.32

P = 9

 

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.

 

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

 

Soln:

 

x f fx
15 3 45
16 8 128
17 10 170
18 10 180
19 5 95
20 4 80
N = 40 Sum = 698

 

Mean age = sum/ N

= 698/ 40

= 17.45 years

 

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 P 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.

 

Soln: Let the number candidates from school III = P

 

Schools No of candidates Ni Average scores (xi)
I 60 75
II 48 80
III P 55
IV 40 50

 

Given

Average score or all schools = 66

\(\frac{N_{1}\overline{x_{1}}+N_{2}\overline{x_{2}}+N_{3}\overline{x_{3}}+ N_{4}\overline{x_{4}}}{N_{1}+ N_{2}+ N_{3}+ N_{4}}\)

\(\frac{4500 + 3340 + 55p + 2000 }{ 60 + 48 + p + 40}\)

10340 + 55p = 66p + 9768

10340 – 9768 = (66 – 55)p

P = 572/11

P = 52

 

10.Five coins were simultaneously tossed 1000 times and at each toss, the number of heads was 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

 

Soln:

 

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

 

No of heads per toss No of tosses fx
0 38 0
1 144 144
2 342 684
3 287 861
4 164 656
5 25 125

 

Mean number of heads per toss = 2470/1000 = 2.47

Mean = 2.47

 

12.The arithmetic mean of the following data is 25. Find the value of k.

 

Xi: 5              15                        25               35           45
fi: 3              k                          3                6              2

 

Sol:

X ­f fx
5 3 15
15 k 15k
25 3 75
35 6 210
45 2 90
N = k + 120 Sum = 15k + 390

Given mean = 25

Sum/ N = 25

15k + 390 = 25k + 350

25k – 15k = 40

10k = 40

k = 4

 

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

 

Xi: 10           15           p          25               30
Fi: 5              10          7            8                 2

 

Soln:

 

X ­f fx
10 5 50
15 10 150
P 7 7p
25 8 200
30 2 60
N = k + 120 Sum = 1p + 460

 

Given mean = 18.75

Sum/ N = 1.75

7p + 460 = 600

7p = 140

P = 20

 

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

                               

Soln:

X f fx
15 2 30
17 3 51
19 4 76
20 + p 5p 100p+5p2
23 6 138
N = 5p +15` Sum = 295 +100p +5p2

 

Given Mean= 2n

Sum/ N = 20

(295 + 100p – 5p2) / (5 + 15) = 20

295 + 100p + 5p2=100p + 300

5p2 – 5 = 0

5(p2 – 1) = 0

P2 – 1 = 0

P = (+1 , -1)

If p + 1 = 0

P = -1

Or p -1 = 0

P = 1

 

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

 

X: 10           30                        50               70           90
f: 17           f1                         32               f2                   19

 

Soln:

 

x f fx
10 17 170
30 f1 30f1
50 32 1600
70 f2 70f2
90 19 1710
N = 120 Sum = 30f1+70f2+3480

 

Given mean

Sum/ N = 50

30f1+70f2+3480/ 120 = 50

30f1+70f2+3480 = 6000 —- (1)

Also, sum of f = 120

17 + f1 + 32 + f2 + 19 = 120

f1 + f2 = 52

f1 = 52 – f2

Substituting the value of f1 in (1)

30 (52 – f2) +70f2 + 3480 = 6000

f2 = 24

Hence f1 = 52 -24 = 28

f1 =28    ; f2 = 24