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13. statistics — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS13. statistics1 Mark
MCQ
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
  • A
    $220$
  • ✓
    $210$
  • C
    $200$
  • D
    $105$
✓

Answer

Correct option: B.
$210$
b
$a_1+a_3=10=a_1+d \Rightarrow 5$

$a_1+a_2+a_3+a_4+a_5+a_6=57$

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

$\Rightarrow a_1+a_6=19$

$\Rightarrow 2 a_1+5 d=19 \text { and } a_1+d=5$

$\Rightarrow a_1=2, d=3$

$\text { Numbers }: 2,5,8,11,14,17$

$\text { Variance }=\sigma^2=\text { mean of squares }-\text { square of mean }$ $=\frac{2^2+5^2+8^2+(11)^2+(14)^2+(17)^2}{6}-\left(\frac{19}{2}\right)^2 ~\\ =\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$

$8 \sigma^2=210$

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