JEE Mains · Maths · STD 11 - 13. statistics
Let \(x_1, x_2, \ldots, x_{10}\) be ten observations such that \(\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta\gt2\), and their variance is \(\frac{4}{5}\). If \(\mu\) and \(\sigma^2\) are respectively the mean and the variance of \(2\left(x_1-1\right)+4 \beta\), \(2\left(x_2-1\right)+4 \beta, \ldots ., 2\left(x_{10}-1\right)+4 \beta\), then \(\frac{\beta \mu}{\sigma^2}\) is equal to :
- A 100
- B 120
- C 110
- D 90
Answer & Solution
Correct Answer
(A) 100
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sum_{l=1}^{10}\left(x_l-2\right)=30 \\ & \sum_{i=1}^{10} x_l=50 \\ & \Rightarrow \text { Mean }=5 \\ & \text { Variance }=\frac{4}{5}=\frac{\sum x_l^2}{10}-(\bar{x})^2 \\ & \frac{4}{5}=\frac{\sum x_l^2}{10}-25 \\ & \Rightarrow \sum x_l^2=258 \end{aligned}\)…
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