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USE CODE: SELF10

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

Mean (x) = 2680/106 = 25


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 + f Sum = 30f+ 70f+ 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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