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Calculate the mean for the following distribution:
= 281/40 = 7.025.
Find the mean of the following data:
= 2650/106 = 25.
The mean of the following data is 20.6 .Find the value of p.
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.
If the mean of the following data is 15, find p.
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.
Find the value of p for the following distribution whose mean is 16.6.
Mean = 16.6
⇒ 24p + 1228 = 1660
⇒ 24p = 1660 − 1228
⇒ 24p = 432
⇒ p = 432/24 = 18
⇒ p = 18
∴ p = 18.
Find the missing value of p for the following distribution whose mean is 12.58.
Mean = 12.58
⇒ ∑fx/N = 12.58
⇒ 7p + 524 = 629
⇒ 7p = 629 − 524
⇒ 7p = 105
⇒ p = 1057 = 15
⇒ p = 15
Find the missing frequency (p) for the following distribution whose mean is 7.68.
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.
Find the value of p, if the mean of the following distribution is 20.
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.
Find the mean of the following distribution:
= 800/40 = 20.
Candidates of four schools appear in a mathematics test. The data were as follows:
If the average score of the candidates of all four schools is 66, Find the number of candidates that appeared from school III.
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.
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.
∴ Mean number of heads per toss = ∑fx/N
= 2470/1000
= 2.47
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.
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
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Chapter 24: Measures of Central Tendency Exercise...