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The mean marks of students of a school is 56. The mean marks of girls is 60 and that of boys is 50. If number of boys and girls are n and m respectively, then <!-- [if gte mso 9]><xml>
The mean marks of students of a school is 56. The mean marks of girls is 60 and that of boys is 50. If number of boys and girls are n and m respectively, then <!-- [if gte mso 9]><xml>
56 n + 56 m = 50 n + 60 m
6n = 4m
Upvote 
Coefficient of variation of two distributions are 50% and 60% and their arithmetic means are 30 and 25 respectively. Difference of standard deviation is:
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Coefficient of variation of two distributions are 50% and 60% and their arithmetic means are 30 and 25 respectively. Difference of standard deviation is:
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C.V. = (σ/x? ) × 100 ⇒ σ = (C.V. × x? ) / 100
∴ σ? = (50×30)/100 = 15 and σ? = (60×25)/100 = 15 ⇒ σ? - σ? = 0
Upvote 
mean = Σx? f? /Σf? = (32+8α+9β)/ (8+α+β)=6
⇒2α+3β=16 . (i)
d? =x? −x? =−4,0,2,3
f? d? ²=64,0,4α,9β
Variance σ²=Σf? d? ²/Σf? =6.8
⇒ (64+4α+9β)/ (8+α+β)=6.8
⇒2.8α+ (−2.2β)=9.6
⇒28α−22β=96
14α−11β=48 . (ii)
Solving (i) and (ii),
⇒β=2, α=5
New mean=Σx? f? /Σf? =85/15=17/3
Upvote If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to
If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to
Var (1, 2, ., n) = (Σn²/n) - (Σn/n)² = 10.
(n (n+1) (2n+1)/6n) - (n (n+1)/2n)² = 10.
(n+1) (2n+1)/6 - (n+1)/2)² = 10.
(n+1)/12 * [2 (2n+1) - 3 (n+1)] = 10.
(n+1)/12 * (4n+2 - 3n-3) = 10.
(n+1) (n-1)/12 = 10.
n² - 1 = 120 ⇒ n² = 121 ⇒ n = 11.
Var (2, 4, ., 2m) = Var (2* (1, 2, ., m) = 2² * Var (1,
Upvote 
1st observation: n? =10, mean x? =2, variance σ? ²=2.
Σx? = n? x? = 20.
σ? ² = (Σx? ² / n? ) - x? ² => 2 = (Σx? ²/10) - 2² => 6 = Σx? ²/10 => Σx? ² = 60.
2nd observation: n? , mean y? =3, variance σ? ²=1. Let n? =n.
Σy? = ny? = 3n.
σ? ² = (Σy? ² / n) - y? ² => 1 = (Σy? ²/n) - 3² => 10 = Σy? ²/n => Σy? ²
Upvote Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?
Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?
Variance of a, b, c & a+2, b+2, c+2, are same.
Given: b = a + c (i)
d² = (1/3) (a² + b² + c²) - [ (a+b+c)/3]²
as a + c = b
d² = (1/3) (a² + b² + c²) - (2b/3)²
9d² = 3 (a² + b² + c²) - 4b²
⇒ b² = 3 (a² + c²) - 9d²
Upvote Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a² + b² is equal to :
Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a² + b² is equal to :
The data consists of n values of a and n values of -a.
Mean x? = (n*a + n* (-a) / 2n = 0 / 2n = 0.
Variance σ² = (Σx? ²)/N - x? ² = (n*a² + n* (-a)²) / 2n - 0² = 2na² / 2n = a².
If a value b is added to all observations, the new mean is x? ' = x? + b = 0 + b = b.
We are given the new mean is 5, so b=5.
A
Upvote The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2,4,10,12,14, then the absolute difference of the remaining two observations is:
The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2,4,10,12,14, then the absolute difference of the remaining two observations is:
x? = (2+4+10+12+14+x+y)/7 = 8
⇒ 42+x+y = 56 ⇒ x+y = 14
σ² = (Σx? ²/n) - (x? )²
16 = (4+16+100+144+196+x²+y²)/7 - (8)²
⇒ 16+64 = (460+x²+y²)/7
⇒ 560 = 460+x²+y² ⇒ x²+y² = 100
⇒ xy=48
(x-y)² = (x+y)² - 4xy = 4
|x-y| = 2
Upvote