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13.Statistics
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ધારોકે છ સંખ્યાઓ $a_1, a_2, a_3, a_4, a_5, a_6$ સમાંતર શ્રેણીમાં છે અને $a_1+a_3=10$. જો આ છ સંખ્યાઓ નું મધ્યક $\frac{19}{2}$ હોય અને તેમનું વિયરણ $\sigma^2$ હોય, તો $8 \sigma^2=........$
A
$220$
B
$210$
C
$200$
D
$105$
(JEE MAIN-2023)
Solution
$a_1+a_3=10=a_1+d \Rightarrow 5$
$a_1+a_2+a_3+a_4+a_5+a_6=57$
$\Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57$
$\Rightarrow a_1+a_6=19$
$\Rightarrow 2 a_1+5 d=19 \text { and } a_1+d=5$
$\Rightarrow a_1=2, d=3$
$\text { Numbers }: 2,5,8,11,14,17$
$\text { Variance }=\sigma^2=\text { mean of squares }-\text { square of mean }$ $=\frac{2^2+5^2+8^2+(11)^2+(14)^2+(17)^2}{6}-\left(\frac{19}{2}\right)^2 ~\\ =\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$
$8 \sigma^2=210$
Standard 11
Mathematics
Similar Questions
જો તો વિચરણ $\sigma^2$ =…………………………..
| $x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
| $f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
