Chapter 24: Measures of Central Tendency Exercise – 24.2
Question: 1
Calculate the mean for the following distribution:
| x: | 5 | 6 | 7 | 8 | 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 | ∑fx = 281 |
= 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 |
| 19 | 13 | 247 |
| 21 | 15 | 315 |
| 23 | 16 | 368 |
| 25 | 18 | 450 |
| 27 | 16 | 432 |
| 29 | 15 | 435 |
| 31 | 13 | 403 |
| N = 106 | ∑fx = 2650 |
= 2650/106 = 25.
Question: 3
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 |
| 15 | 10 | 150 |
| P | 25 | 25p |
| 25 | 7 | 175 |
| 35 | 5 | 175 |
| N = 50 | ∑fx = 25p + 530 |
It is given that,
Mean = 20.6
⇒ 25p + 530 = 20.6 × 50
⇒ 25p = 1030 − 530
⇒ 25p = 500
⇒ p = 500/25 = 20
⇒ p = 20
∴ p = 20.
Question: 4
If the mean of the following data is 15, find p.
| f: | 6 | p | 6 | 10 | 5 |
| x: | 5 | 10 | 15 | 20 | 25 |
Solution:
| x | f | fx |
| 5 | 6 | 30 |
| 10 | P | 10p |
| 15 | 6 | 90 |
| 20 | 10 | 200 |
| 25 | 5 | 125 |
| N = p + 27 | ∑fx = 10p + 445 |
It is given that,
Mean = 15
⇒∑fx/N = 15
⇒10p + 445p + 27=15
⇒10p + 445 = 15 × (p + 27)
⇒ 10p + 445 = 15p + 405
⇒15p − 10p = 445 − 405
⇒ 5p = 40
⇒ p = 405 = 8
⇒ p = 8
∴ p = 8.
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 | 16 | 192 |
| 15 | 20 | 300 |
| P | 24 | 24p |
| 20 | 16 | 320 |
| 25 | 8 | 200 |
| 30 | 4 | 120 |
| N = 100 | ∑fx = 24p + 1228 |
It is given that,
Mean = 16.6
⇒ 24p + 1228 = 1660
⇒ 24p = 1660 − 1228
⇒ 24p = 432
⇒ p = 432/24 = 18
⇒ p = 18
∴ 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 | 4 | 80 |
| 25 | 2 | 50 |
| N = 50 | ∑fx = 7p + 524 |
It is given that,
Mean = 12.58
⇒ ∑fx/N = 12.58
⇒ 7p + 524 = 629
⇒ 7p = 629 − 524
⇒ 7p = 105
⇒ p = 1057 = 15
⇒ p = 15
∴ p = 18.
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 | ∑fx = 9p + 303 |
It is given that,
Mean = 7.68
⇒ ∑fx/N = 7.68
⇒ 9p + 303p + 41 = 7.68
⇒ 9p + 303 = 7.68p + 314.88
⇒ 9p − 7.68p = 314.88 − 303
⇒ 1.32p = 11.88
⇒ p = 11.881.32 = 9
⇒ p = 9
∴ p = 9.
Question: 8
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 | Fx = 5p2 + 100p + 295 |
It is given that,
Mean = 20
⇒ 5p2 + 100p + 295 = 20(5p + 15)
⇒ 5p2 + 100p + 295 = 100p + 300
⇒ 5p2 = 300 − 295
⇒ 5p2 = 5
⇒ p2 = 1
⇒ p = ±1
Frequency can’t be negative.
Hence, value of p is 1.
Question: 9
Find the mean of the following distribution:
| x: | 10 | 12 | 20 | 25 | 35 |
| f: | 3 | 10 | 15 | 7 | 5 |
Solution:
| x | f | fx |
| 10 | 3 | 30 |
| 12 | 10 | 120 |
| 20 | 15 | 300 |
| 25 | 7 | 175 |
| 35 | 5 | 175 |
| N = 40 | ∑fx = 800 |
= 800/40 = 20.
Question: 10
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 four schools is 66, Find the number of candidates that appeared from school III.
Solution:
| Schools | No. of Candidates | Average Score |
| I | 60 | 75 |
| II | 48 | 80 |
| III | x | 55 |
| IV | 40 | 50 |
Given the average score of all schools = 66
⇒ 10340 + 55x = 66x + 9768
⇒ 10340 − 9768 = 66x − 55x
⇒ 11x = 572
⇒ x = 572/11 = 52
∴ No. of candidates appeared from school III = 52.
Question: 11
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 (x) | No. of tosses (f) | fx |
| 0 | 38 | 0 |
| 1 | 144 | 144 |
| 2 | 342 | 684 |
| 3 | 287 | 861 |
| 4 | 164 | 656 |
| 5 | 25 | 125 |
| N = 1000 | ∑fx = 2470 |
∴ Mean number of heads per toss = ∑fx/N
= 2470/1000
= 2.47
Question: 12
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 |
Solution:
| x | f | fx |
| 10 | 17 | 170 |
| 30 | f1 | 30f1 |
| 50 | 32 | 1600 |
| 70 | f2 | 70f2 |
| 90 | 19 | 1710 |
| N = 120 | ∑fx = 3480 + 30f1 + 70f2 |
It is given that
Mean = 50
⇒ ∑fx/N = 50
⇒ 3480 + 30f1 + 70f2 = 50 × 120
⇒30f1 + 70f2 = 6000 − 3480
⇒10(3f1 + 7f2) = 10(252)
⇒ 3f1 + 7f2 = 252 ⋅⋅⋅⋅ (1) [∵ Divide by 10]
And N = 20
⇒ 17 + f1 + 32 + f2 + 19 = 120
⇒ 68 + f1 + f2 = 120
⇒ f1 + f2 = 120 − 68
⇒ f1 + f2 = 52
Multiply with 3 on both sides
⇒ 3f1 + 3f2 = 156 ⋅⋅⋅ (2)
Subtracting equation (2) from equation (1)
⇒ 3f1 + 7f2 − 3f1 − 3f2 = 252 − 156
⇒ 4f2 = 96
⇒ f2 = 96/4 = 24
Put the value of f2 in equation (1)
⇒ 3f1 + 7 × 24 = 252
⇒ 3f1 = 252 − 168
⇒ f1 = 84/3 = 28
⇒ f1 = 28