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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Statistics

Let (x_1, x_2, , x_ 10 ) be ten observations such that (∑_ i=1 ^ 10...

Let (x_1, x_2, \ldots, x_{10}) be ten observations such that (\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98,2=98,\beta\gt2),andtheirvarianceis(2), and their variance is (\frac{4}{5}.If(μ). If (\mu) and (\sigma^2)arerespectivelythemeanandthevarianceof(2(x1−1)+42) are respectively the mean and the variance of (2 \left(x_1-1\right)+4\beta,(2(x2−1)+4, (2 \left(x_2-1\right)+4\beta, \ldots .,2(x10−1)+42\left(x_{10}-1\right)+4\beta,then(βμσ2i, then (\frac{\beta \mu}{\sigma^2} is equalto:to :

Options

  1. A100
  2. B120
  3. C110
  4. D90

Correct answer

A. 100

Step-by-step solution

∑l=110(xl−2)=30∑i=110xl=50⇒ Mean =5 Variance =45=∑xl210−(xˉ)245=∑xl210−25⇒∑xl2=258\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, ∑l=110(xl−β)2=98\sum_{l=1}^{10}\left(x_l-\beta\right)^2=98∑l=110xl2−2β∑l=110xl+10β2=98⇒258−2β(50)+10β2=98⇒10β2−100β+160=0⇒β2−10β+16=0⇒β=8 as β>2\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, as per the question $2\left(x_1-1\right)+4 \beta, 2\left(x_2-1\right)+4 \beta, \ldots 2\left(x_{10}-1\right)+4 \beta$ Can be simplified as $2 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}$$$
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