Jee Main 2024 29 01 2024 Shift 2 - Question10
Question 10
The mean of 5 observations is $\frac{24}{5}$ and variance is $\frac{194}{25}$. If the mean of first four observations is $\frac{7}{2}$, then the variance of first four observations is
(1) $\frac{3}{2}$
(2) $\frac{5}{2}$
(3) $\frac{5}{4}$
(4) $\frac{2}{3}$
Show Answer
Answer: (3)
Solution:
$\sum _{i=1}^{5} x _i=24$
$$ \begin{aligned} & \frac{\sum x _i^{2}}{5}-\left(\frac{24}{5}\right)^{2}=\frac{194}{25} \\ & \Rightarrow \quad \sum x _i^{2}=\frac{770}{25} \times 5=154 \end{aligned} $$
$5^{\text {th }}$ observation $=24-\frac{7}{2} \times 4=10$
$$ \begin{aligned} \text { New variance } & =\frac{\sum _{i=1}^{4} x _i^{2}}{4}-\left(\frac{7}{2}\right)^{2} \\ & =\frac{154-100}{4}-\frac{49}{4} \\ & =\frac{5}{4} \end{aligned} $$
BETA
