Statistics Ans 1
Q1 - 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
$=\frac{2^{2}+5^{2}+8^{2}+(11)^{2}+(14)^{2}+(17)^{2}}{6}-(\frac{19}{2})^{2}$
$=\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$
$\therefore 8 \sigma^{2}=210$
