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QJEEM23Ma1d4c: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1d4c Solution: Let the mean of the data .tg…

Let the mean of the data

xx 1 3 5 7 9
Frequency ( ff ) 4 24 28 α\alpha 8

be 5. If mm and σ2\sigma^{2} are respectively the mean deviation about the mean and the variance of the data, then 3αm+σ2\frac{3 \alpha}{m+\sigma^{2}} is equal to __________

Step-by-Step Verified Solution

amp;5=xˉ=∑xifi∑fi=4+72+140+7α+7264+αamp;⇒320+5α=288+7α⇒2α=32⇒α=16\begin{aligned}& 5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha} \\\\& \Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16\end{aligned}

∑fiamp;=80 M.D amp;=∑fi∣xi−5∣∑fiamp;=4+4+24×2+0+16×2+8×480amp;=85\begin{aligned}\sum f_i & =80 \\\\\text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\& =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\\\& =\frac{8}{5}\end{aligned}

σ2amp;=∑fi∣xi−5∣∑fiamp;=4+16+24×4+0+16×4+8×1680=225\begin{aligned}\sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\& =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5}\end{aligned}

So, 3αm+σ2=3×1685+225=8\frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8

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