The variance of the first 20 positive integral multiples of 4 is equal to

Question

The variance of the first 20 positive integral multiples of 4 is equal to (A) 532 (B) 133 (C) 266 (D) 600

Tardigrade Admin · Accepted Answer

The numbers are 4,8,12........80 Mean, (. barx.)=(4 (1 + 2 + 3 + . . . . . . 20)/20)=42 Sigma xi2=42(12 + 22 + (. . . . . . 20)2) =(16 × 20 × 21 × 41/6)=45920 So, the variance is =( Sigma xi2/20)-(. overset-x .)2=2296-1764=532

Q. The variance of the first $20$ positive integral multiples of $4$ is equal to

$532$

$133$

$266$

$600$

The numbers are $4, 8, 12........80$
Mean, $(\overset{x}{ˉ}) = \frac{4 ( 1 + 2 + 3 + ......20 )}{20} = 42$
$Σ x_{i}^{2} = 4^{2} (1^{2} + 2^{2} + (......20)^{2})$
$= \frac{16 \times 20 \times 21 \times 41}{6} = 45920$
So, the variance is $= \frac{Σ x _{i}^{2}}{20} - (x -)^{2} = 2296 - 1764 = 532$

Q. The variance of the first 20 positive integral multiples of 4 is equal to

532

133

266

600