$\begin{aligned}
& \sum_{l=1}^{10}\left(x_l-2\right)=30 \
& \sum_{i=1}^{10} x_l=50 \
& \Rightarrow \text { Mean }=5 \
& \text { Variance }=\frac{4}{5}=\frac{\sum x_l^2}{10}-(\bar{x})^2 \
& \frac{4}{5}=\frac{\sum x_l^2}{10}-25 \
& \Rightarrow \sum x_l^2=258
\end{aligned}Now,\sum_{l=1}^{10}\left(x_l-\beta\right)^2=98\begin{aligned}
& \sum_{l=1}^{10} x_l^2-2 \beta \sum_{l=1}^{10} x_l+10 \beta^2=98 \
& \Rightarrow 258-2 \beta(50)+10 \beta^2=98 \
& \Rightarrow 10 \beta^2-100 \beta+160=0 \
& \Rightarrow \beta^2-10 \beta+16=0 \
& \Rightarrow \beta=8 \text { as } \beta>2
\end{aligned}Now,asperthequestion2\left(x_1-1\right)+4 \beta, 2\left(x_2-1\right)+4 \beta, \ldots 2\left(x_{10}-1\right)+4 \betaCanbesimplifiedas2 x_1+30,2 x_2+30, \ldots, 2 x_{10}+30\begin{aligned} & \mu=2(5)+30=40 \ & \sigma^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5} \ & \frac{\beta \mu}{\sigma^2}=\frac{8 \times 40}{\frac{16}{5}}=100\end{aligned}$
Let x_1, x_2, …, x_10 be ten observations such that Σ_i=1^10 (x_i-2 )=30, Σ_i=1^10 (x_i-β )^2=98, βgt2, and their variance is 4 5. If μ and σ^2 are…
JEE Main 2025 — Mathematics Probability & Statistics
Let x1,x2,…,x10 be ten observations such that i=1∑10(xi−2)=30,i=1∑10(xi−β)2=98,β>2, and their variance is 54. If μ and σ2 are respectively the mean and the variance of 2(x1−1)+4β, 2(x2−1)+4β,….,2(x10−1)+4β, then σ2βμ is equal to :
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
100
- B
120
- C
110
- D
90
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