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Statistics Part 2

Section: Mathematics

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Question : 21 of 50

Marks: +1, -0

Let

x_1, x_2, \ldots, x_{10}

be ten observations such that

\sum\limits_{i=1}^{10} (x_i-2)=30, \;\; \sum\limits_{i=1}^{10} (x_i-\beta)^2=98, \beta>2

, and their variance is

\;\frac{4}{5}

\mu

\sigma^2

are respectively the mean and the variance of

2 (x_1-1)+4\beta, 2 (x_2-1)+4\beta, \ldots

2 (x_{10}-1)+4\beta

\;\frac{\beta \mu}{\sigma^2}

[29 Jan 2025 Shift 1]

110
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120

Solution: 👈: Video Solution

\;\sum\limits_{i=1}^{10} (x_i-2)=30

\;\sum\limits_{i=1}^{10} x_i=50

\;\Rightarrow \;\text{ Mean }\;=5

\;\;\text{ Variance }\;=\;\frac{4}{5}=\;\frac{\Sigma x_i^2}{10}-(\overline{x})^2

\;\;\frac{4}{5}=\;\frac{\Sigma x_i^2}{10}-25

\;\Rightarrow \sum x_i^2=258

Now,

\sum\limits_{i=1}^{10} (x_i-\beta)^2=98

\;\sum\limits_{i=1}^{10} x_i^2-2\beta \sum\limits_{i=1}^{10} x_i+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

Now, as per the question

2 (x_1-1)+4\beta, 2 (x_2-1)+4\beta, \ldots, 2 (x_{10}-1)+4\beta

Can be simplified as

2x_1+30, 2x_2+30, \ldots, 2x_{10}+30

\;\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

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