Test: Variance And Standard Deviation - JEE MCQ

Variance is the mean of the squares of the deviations from the mean. Variance is the square of standard deviation. Therefore any unit of a given set is converted into squares at the time of calculating the variance.

sum of weights of 10 items = 280
sum of weights of n items = 35n
so, sum of weights of (10 + n) items = 280 + 35n
so ,mean = (280 + 35 n) / (10 + n)
30(10 + n) = 280 + 35n 
solving we get, n = 4

If a three, is added to each number in a set of data, the mean will be increased by 3 and the standard deviation will be unaltered (since the spread of the data will be unchanged).
Hence, variance of the new data = 20

Variance = 15 
New variance = 2^2*15 
= 4*15 
= 60

Answer:  C
Solution: Variance= [summation (y^2×f) /N] -[ summation (yf) /N]^2
=(296/25) -(0/25) ^2
=11.84
standard deviation=√11.84=3.12

First 10 multiples of 4 are 4,8,12...40.
This is an A.P.
sum=n/2(a+l)
 = 10/2(4+40)
∴ sum=220.
Mean, u=sum/n
= 220/10 = 22
D1 = 4-22 = -18
D2 = 8-22 = -16
D3 = 12-22 = -10
D4 = 16- 22 = -8
Similarly we subtract multiple of 4 by 22 upto 10 terms we get 
-18, -14, -10, -8………...18
S.D. = σ² = ∑(D²)/n
        =[ (-18)² ,(-14)², (-10)², (-6)² + (-2)² +(6)² + (10)² + (14)² + (18)²]/10
Solving this, we get
σ = 11.5​