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# Chapter 7: Statistics Exercise – 7.1

### Question: 1

Calculate the mean for the following distribution:

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

### Solution:

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

Mean = 281/40 =7.025

### 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:

 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

### Question: 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

### Solution:

 x f fx 10 3 30 5 10 150 P 25 25p 25 7 175 35 5 175 N = 50 Sum = 2620 + 25P

Given Mean = 20.6

20.6 = (530 + 25p)/50

20.6 25p = 20.6 P

P = 20

### Question: 4

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

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

### Solution:

 x f fx 5 6 30 10 P 10p 15 6 90 20 10 200 25 5 125 N = p + 27 Sum = 10p + 445

Given Mean =15 15P + 405 = 10P + 445

15P - 10P = 445-405

5P = 40

P = 5

### 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:

 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

16.6 = (24p +1228)/100

1660  =  24p + 1228

24p = 1660 – 1228

P = 432/24

P = 18

### Question: 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

### Solution:

 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
Mean = Sum/N
12.58 =  (524 +7p)/50
629 = 524 + 7p
629 - 524 = 7p
105 = 7P
P = 105/7
P = 15

### 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:

 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 7.68 (P + 41) = 9P + 303
7.68 P + 314.88 = 9P + 303
9P – 7.68 P = 314.88 – 303
1.32 P = 11.88
P = 11.88/1.32
P = 9

### 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.

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

### Solution:

 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

### 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 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.

### Solution:

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 10340 + 55p = 66 p + 9768
10340 - 9768 = 66p –55 p
P = 572/11
P = 52

### Question: 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

### Solution:

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

### Question: 11

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

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

### Solution:

 x f fx 5 3 15 15 k 15k 25 3 75 35 6 210 45 2 90 N = k + 14 Sum = 15k + 390

Given mean = 25
Sum/ N = 25 25K + 350 = 15K + 390
25 K – 15K = 390 - 350
10 k = 40
K = 40/10
K = 4

### Question: 12

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:

 x f fx 10 5 50 15 10 150 20 p 20p 25 8 200 30 2 60 N = P + 25 Sum = 20p + 460

Given mean = 18.75 18.75p + 468.75 = 20p + 460
468.75 – 460 = 20p – 18.75p
8.75 = 1.25P
P = 8.75/1.25
P = 7

### Question: 13

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:

 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 100p + 300 = 295 + 100p + 5p2
300 – 295 = 5p2 + 100p
5 = 5p2 + 100P
5p2 + 100P – 5 =0
5p2 + 20p –5P
5(p2 – 1)  = 0
P2 – 1 = 0
P (+1, -1)
If p + 1 = 0, p = -1, Or p – 1 = 0, P = 1

### Question: 14

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50. If ∑f = 120

 x 10 30 50 70 90 f 17 f1 32 f2 19

### Solution:

 x f fx 10 17 170 30 f1 30f1 50 32 1600 70 f2 70f2 90 19 1710 N = 68 + f1 + f2 Sum = 30f1 + 70f2 + 3480

Given mean = Sum/N
= 50 3400 + 50 f1 + 50 f2 = 30 f1 + 70 f2 + 3480
50 f1 – 30 f1 + 50 f2 – 70 f2 = 3480 – 3400
20 f1 – 20 f2 = 80
f1 – f2 = 4    ….. (i)
f1 + f2 + 68 = 120
f1 + f2 = 52 …. (ii)
Add both equation
f1 = 28
f2 = 24

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