Statistics Ans 10
Q10 - 01 February - Shift 2
Let $9=x_1<x_2<\ldots<x_7$ be in an A.P. with common difference $d$. If the standard deviation of $x_1, x_2 \ldots$, $x_7$ is 4 and the mean is $\overline{x}$, then $\overline{x}+x_6$ is equal to :
(1) $18(1+\frac{1}{\sqrt{3}})$
(2) 34
(3) $2(9+\frac{8}{\sqrt{7}})$
(4) 25
Show Answer
Answer: (2)
Solution:
Formula: Arithmetic Progression, Arithmetic mean of individual series (ungrouped data), Variance of individual observations (ungrouped data), Common difference of an A.P (v)
$ x_1=9, \quad x_2=9+d, \quad x_3=9+2 d, \quad x_4=9+3 d, \quad x_5=9+4 d, \quad x_6=9+5 d, \quad x_7=9+6 d $
Mean $\mathrm{x}=\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4+\mathrm{x}_5+\mathrm{x}_6+\mathrm{x}_7}{7}$
$ \begin{aligned} & =\frac{63+21 \mathrm{~d}}{7}=9+3 \mathrm{~d} \ &\text{Variance,}=\sum_{\mathrm{i=1}}^7\frac{\mathrm{x}-\mathrm{x}_{\mathrm{i}}^2}{\mathrm{~N}} \ & =\frac{29 \mathrm{~d}^2+4 \mathrm{~d}^2+\mathrm{d}^2}{7}=4 \mathrm{~d}^2 \end{aligned} $
Given standard deviation $\sigma=4$ Variance $\sigma^2=16$
$ 4 \mathrm{~d}^2=16 \Rightarrow \mathrm{d}=2 $
Mean $x=9+3(2)=15$
$ \mathrm{x}_6=9+5 \mathrm{~d}=19 $
$ \therefore x+\mathrm{x}_6=15+19=34 $
