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QJEEM23Ma9a0e: JEE Main 2023 Mathematics - Class 11 JEEM23Ma9a0e Solution: If the mean of the frequency distribution .tg…

If the mean of the frequency distribution

Class : 0-10 10-20 20-30 30-40 40-50
Frequency : 2 3 xx 5 4

is 28, then its variance is __________.

Step-by-Step Verified Solution

Given mean is 28

 So, 2×5+3×15+x×25+5×35+4×4514+x=28⇒10+45+25x+175+18014+x=28⇒310+25x=392+28x⇒3x=18⇒x=6\begin{array}{ll}\text { So, } \frac{2 \times 5+3 \times 15+x \times 25+5 \times 35+4 \times 45}{14+x}=28 \\\\\Rightarrow \frac{10+45+25 x+175+180}{14+x}=28 \\\\\Rightarrow 310+25 x=392+28 x \\\\\Rightarrow 3 x=18 \Rightarrow x=6\end{array}

amp;∴ Variance =(∑xi2fi∑fi)−( mean )2amp;=(2×52+3×152+6×252+5×352+4×45220)−(28)2amp;=(50+675+3750+6125+810020)−(28)2amp;=(1870020)−(28)2amp;=935−784=151\begin{aligned}& \therefore \text { Variance }=\left(\frac{\sum x_i^2 f_i}{\sum f_i}\right)-(\text { mean })^2 \\\\& =\left(\frac{2 \times 5^2+3 \times 15^2+6 \times 25^2+5 \times 35^2+4 \times 45^2}{20}\right)-(28)^2 \\\\& =\left(\frac{50+675+3750+6125+8100}{20}\right)-(28)^2 \\\\& =\left(\frac{18700}{20}\right)-(28)^2 \\\\& =935-784=151\end{aligned}

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