First slide

Introduction to statistics

Question

$The mean and variance of ' n'observations x_{1}, x_{2}, x_{3}, \dots, x_{n} are 6 and 0, respectively. If$ $\sum_{i = 1}^{n} x_{i}^{2} = 900, then the value of n is equal to$

Moderate

A

80
B

25
C

20
D

16

Solution

$\begin{array}{l} Given mean \bar{x} = 6 and \\ variance σ^{2} = \frac{\sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}}{n} = 0 \\ \Rightarrow {(x_{1} - 6)}^{2} + {(x_{2} - 6)}^{2} + \dots \dots + {(x_{n} - 6)}^{2} = 0 \\ \Rightarrow x_{1} = x_{2} = \dots \dots . . = x_{n} = 6 \end{array}$
$(∵ Sum of perfect squares will be zero, when each of them is zero)$
$\begin{aligned} Given that \sum_{i = 1}^{n} x_{i}^{2} = 900 \\ \Rightarrow n \cdot 6^{2} = 900 \\ ∴ n = 25 \end{aligned}$

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