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Statistics Part 1

Section: Mathematics

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Question : 18 of 100

Marks: +1, -0

Let the mean and the variance of 5 observations

x_1, x_2, x_3, x_4, x_5

\;\frac{24}{5}

\;\frac{194}{25}

respectively. If the mean and variance of the first 4 observation are

\;\frac{7}{2}

and a respectively, then

(4 a+x_5)

[29-Jun-2022-Shift-1]

13
15
17
18

Solution: 👈: Video Solution

Mean

(\overline{x}) = \;\frac{x_1+x_2+x_3+x_4+x_5}{5}

Given,

\;\frac{x_1+x_2+x_3+x_4+x_5}{5} = \;\frac{24}{5}

\Rightarrow x_1+x_2+x_3+x_4+x_5=24

Now, Mean of first 4 observation

= \;\frac{x_1+x_2+x_3+x_4}{4}

Given,

= \;\frac{x_1+x_2+x_3+x_4}{4} = \;\frac{7}{2}

\Rightarrow x_1+x_2+x_3+x_4=14

From equation (1) and (2), we get

14+x_5=24

\Rightarrow x_5=10

Now, variance of first 5 observation

= \;\frac{\text{sumx}_i^2}{n} - (\overline{x})^2

= \;\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5} - \left(\;\frac{24}{5}\right)^2

Given,

\;\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5} - \left(\;\frac{24}{5}\right)^2 = \;\frac{194}{24}

\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2 = 5\left(\;\frac{194}{25}+\;\frac{576}{25}\right)

\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=154

\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+(10)^2=154

\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2=54

Now, variance of first 4 observation

= \;\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4} - \left(\;\frac{7}{2}\right)^2

Given,

\;\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4} - \left(\;\frac{7}{2}\right)^2 = a

Given,

\;\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4} - \left(\;\frac{7}{2}\right)^2 = a

\Rightarrow \;\frac{54}{4} - \;\frac{49}{4} = a

\Rightarrow a = \;\frac{5}{4}

\therefore 4a+x_5

= 4 \times \;\frac{5}{4}+10=15

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