Statistics Question 1

Question 1 - 24 January - Shift 2

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in A.P. and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^{2}$, then $8 \sigma^{2}$ is equal to

(1) 220

(2) 210

(3) 200

(4) 105

Show Answer

Answer: (2)

Solution:

Formula: Arithmetic Progression, Arithmetic mean of individual series (ungrouped data), Variance of individual observations (ungrouped data)

$a_1+a_3=10=a_1+d \Rightarrow 5$

$\mathbf{a} _1+\mathbf{a} _2+\mathbf{a} _3+\mathbf{a} _4+\mathbf{a} _5+\mathbf{a} _6=\mathbf{5 7}$

$\Rightarrow \frac{6}{2}[a_1+a_6]=57$

$\Rightarrow a_1+a_6=19$

$\Rightarrow 2 a_1+5 d=19$ and $a_1+d=5$

$\Rightarrow a_1=2, d=3$

Numbers : $2,5,8,11,14,17$

Variance $=\sigma^{2}=$ mean of squares - square of mean

$=\frac{2^{2}+5^{2}+8^{2}+(11)^{2}+(14)^{2}+(17)^{2}}{6}-(\frac{19}{2})^{2}$

$=\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$

$\therefore 8 \sigma^{2}=210$