Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Chapter 24: Measures of Central Tendency Exercise – 24.2 Question: 1 Calculate the mean for the following distribution:
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
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Chapter 24: Measures of Central Tendency Exercise...