JEE Main 2025 — Statistics Question with Solution
JEE Main 2025 (7 Apr Shift 1)
Question
The mean and standard deviation of 100 observations are 40 and 5.1 , respectively, By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are and respectively, then is equal to
Choose an option
Show full solutionCorrect option: D
Correct answer
D449
Step-by-step explanation
$\begin{aligned}
& \text { Actual means }=\mu=\frac{100(40)-50+40}{100} \\ & \mu=40-\frac{1}{10}=39.9
\end{aligned}\begin{aligned} & (5.1)^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{100}-(\overline{\mathrm{x}})^2 \\ & \sum \mathrm{x}_{\mathrm{i}}^2=100 \times\left(40^2\right)+100(5.1)^2 \\ & \sum \mathrm{x}_{\mathrm{i}}^2=16 \times 10^4+(5.1)^2 \times 100=162601 \\ & \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2-50^2+40^2}{100}-(\mu)^2 \\ & \sigma^2=1617.01-(39.9)^2=25 \\ & \sigma=5 \\ & 10(\mu+\sigma)=10(39.9+5) \\ & =10 \times 44.9=449\end{aligned}$
& \text { Actual means }=\mu=\frac{100(40)-50+40}{100} \\ & \mu=40-\frac{1}{10}=39.9
\end{aligned}\begin{aligned} & (5.1)^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{100}-(\overline{\mathrm{x}})^2 \\ & \sum \mathrm{x}_{\mathrm{i}}^2=100 \times\left(40^2\right)+100(5.1)^2 \\ & \sum \mathrm{x}_{\mathrm{i}}^2=16 \times 10^4+(5.1)^2 \times 100=162601 \\ & \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2-50^2+40^2}{100}-(\mu)^2 \\ & \sigma^2=1617.01-(39.9)^2=25 \\ & \sigma=5 \\ & 10(\mu+\sigma)=10(39.9+5) \\ & =10 \times 44.9=449\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Statistics chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.