JEE 2024 — Mathematics PYQ

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Question

JEE 2024 — Mathematics PYQ

JEE | Mathematics | 2024

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively,of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to

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Explanation

$\overset{ˉ}{X}_{old} = \frac{24}{5}$

$\Rightarrow \frac{a _{1} + a _{2} + a _{3} + a _{4} + a _{5}}{5} = \frac{24}{5}$

$\Rightarrow \sum_{i = 1}^{5} a_{i} = 24 ... (i)$

$(Var)_{old} = \frac{194}{25}$

$n = 5$

$\overset{ˉ}{X}_{new} = \frac{7}{2}$

$\Rightarrow \frac{\sum _{i = 1}^{4} a _{i}}{4} = \frac{7}{2}$

$\Rightarrow \sum_{i = 1}^{4} a_{i} = 14 ... (ii)$

$a_{5} = 10$

$(Var)_{o / d} = \frac{194}{25}$

$\frac{\sum x _{c}^{2}}{5} - (\overset{ˉ}{X})^{2} = \frac{194}{25}$

$\frac{\sum _{i = 1}^{4} a _{i}^{2} + 100}{5} - (\frac{24}{5})^{2} = \frac{194}{25}$

$\frac{\sum _{i = 1}^{4} a _{i}^{2} + 100}{5} = \frac{194}{25} + \frac{576}{25}$

$\sum_{i = 1}^{4} a_{i}^{2} = \frac{770}{5} - 100 = \frac{270}{5}$

$\sum_{i = 1}^{4} a_{i}^{2} = 54$

$New Variance$

$\frac{\sum _{i = 1}^{4} a _{i}^{2}}{4} - (\overset{ˉ}{X})^{2} = \frac{54}{4} - (\frac{7}{2})^{2} = \frac{5}{4}$

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