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QJEEM23Mad97b: JEE Main 2023 Mathematics - Class 11 JEEM23Mad97b Solution: If the mean and variance of the frequency distribution .tg…

If the mean and variance of the frequency distribution

xix_i 2 4 6 8 10 12 14 16
fif_i 4 4 α\alpha 15 8 β\beta 4 5

are 9 and 15.08 respectively, then the value of α2+β2−αβ\alpha^2+\beta^2-\alpha\beta is ___________.

Step-by-Step Verified Solution

xiamp;fiamp;xi2amp;fixiamp;fixi22amp;4amp;4amp;8amp;164amp;4amp;16amp;16amp;646amp;αamp;36amp;6αamp;36α8amp;15amp;64amp;120amp;96010amp;8amp;100amp;80amp;80012amp;βamp;144amp;12βamp;144β14amp;4amp;196amp;56amp;78416amp;5amp;256amp;80amp;1280\begin{array}{lllll}\hline x_i & f_i & x_i^2 & f_i x_i & f_i x_i^2 \\\hline 2 & 4 & 4 & 8 & 16 \\\hline 4 & 4 & 16 & 16 & 64 \\\hline 6 & \alpha & 36 & 6 \alpha & 36 \alpha \\\hline 8 & 15 & 64 & 120 & 960 \\\hline 10 & 8 & 100 & 80 & 800 \\\hline 12 & \beta & 144 & 12 \beta & 144 \beta \\\hline 14 & 4 & 196 & 56 & 784 \\\hline 16 & 5 & 256 & 80 & 1280 \\\hline\end{array}

∴\therefore amp;Σfi=40+α+β\begin{aligned} & \Sigma f_i=40 +\alpha+\beta\end{aligned}

amp;Σfixi=360+6α+12β\begin{aligned} & \Sigma f_i x_i=360+ 6 \alpha+12 \beta\end{aligned}

amp;Σfixi2=3904+36α+144β\begin{aligned} & \Sigma f_i x_i^2=3904 +36 \alpha+144 \beta\end{aligned}

Given, mean =9=9

amp;⇒ΣfixiΣfi=9amp;⇒360+6α+12β40+α+β=9amp;⇒360+6α+12β=9(40+α+β)amp;⇒3β=3αamp;⇒α=β.........(i)\begin{aligned}& \Rightarrow \frac{\Sigma f_i x_i}{\Sigma f_i}=9 \\\\& \Rightarrow \frac{360+6 \alpha+12 \beta}{40+\alpha+\beta}=9 \\\\& \Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta) \\\\& \Rightarrow 3 \beta=3 \alpha \\\\& \Rightarrow \alpha=\beta .........(i)\end{aligned}

amp; Variance =15.08amp;⇒Σfixi2Σfi−(ΣfixiΣfi)2=15.08amp;⇒3904+36α+144β40+α+β−(9)2=15.08amp;⇒3904+180α40+2α=81+15.08[∵α=β]amp;⇒3904+180α=96.08(40+2α)amp;⇒3904+180α=3843.2+192.16αamp;⇒60.8=12.16αamp;⇒α=5=βamp;∴α2+β2−αβ=25+25−25=25\begin{aligned} & \text { Variance }=15.08 \\\\ & \Rightarrow \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2=15.08 \\\\ & \Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(9)^2=15.08 \\\\ & \Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}=81+15.08 \quad[\because \alpha=\beta] \\\\ & \Rightarrow 3904+180 \alpha=96.08(40+2 \alpha) \\\\ & \Rightarrow 3904+180 \alpha=3843.2+192.16 \alpha \\\\ & \Rightarrow 60.8=12.16 \alpha \\\\ & \Rightarrow \alpha=5=\beta \\\\ & \therefore \alpha^2+\beta^2-\alpha \beta=25+25-25=25\end{aligned}

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