# RD Sharma Solutions Class 10 Statistics Exercise 7.1

### RD Sharma Class 10 Solutions Chapter 7 Ex 7.1 PDF Free 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

Using formula, Mean = fx/N

Mean = 281/40 =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 19 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 = 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 = 50 Sum = 530+25p

Mean = 20.6 (given)

Using formula, Mean = fx/N = (530+25p)/50

Now, (530+25p)/50 = 20.6

p = 20

The value of p is 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 = p+27 Sum = 10p + 445

Mean =15 (Given)

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

10p + 445 = 15p + 405

5p = 40

Or p = 8

The value of p is 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 16 192 15 20 300 p 24 24p 20 16 320 25 8 200 30 4 120 N = 100 Sum = 24p + 1228

Mean = 16.6 (given)

(24p+1228)/100 = 16.6

24p + 1228 = 1660

24p = 1660 – 1228

p = 432/24 = 18

The value of p is 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 4 480 25 2 50 N = 50 Sum = 524 + 7p

Mean = 12.58 (given)

Sum/N = 12.58

(524 +7p)/50 = 12.58

524 + 7p = 629

7p = 105 = 15

The value of p is 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

Mean = 7.68 (given)

(9p+303)/p+41 = 7.68

9p + 303 = (7.68)(p+41)

9p – p(7.68) = 314.88 – 303

1.32p = 11.88

p = (11.88)/1.32 = 9

The value of p is 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 = fx/ 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:

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

Average score or all schools = 66 (given)

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

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

10340 + 55p = 66p + 9768

10340 – 9768 = (66 – 55)p

p = 572/11 =52

The value of p is 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(x) No of tosses(f) 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

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 = 7p + 460

Given : Mean = 18.75

Sum/ N = 18.75

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= 20

Sum/ N = 20

(295 + 100p – 5p2) / (5p + 15) = 20

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

5p2 – 5 = 0

5(p2 – 1) = 0

p = 1 or 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

Sum of f is 120.

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 = 50

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 and f2 = 24