Answer to Question #283459 in Statistics and Probability for Md Altaf Raja

Question

The random variable X is normally distributed with mean 68 cm and 2.5 cm. What should be the size of the sample whose

mean shall not differ from the population mean by more than 1 cm with probability 0.95? (Given that area under standard

normal curve to the right of the ordinate at 1.96 is 0.025)

Answer 1

Solution:

$X\sim N(\mu,\sigma) \\ \mu=68,\sigma=2.5$

$\bar X\sim N(\mu_X,\sigma_X) \\ \mu_X=\mu=68,\sigma_X=\sigma/\sqrt n=2.5/\sqrt n$

$P(67\le \bar X\le 69)=0.95 \\ \Rightarrow P(\bar X\le 69)-P( \bar X\le 67)=0.95 \\ \Rightarrow P(z\le \dfrac{69-68}{2.5/\sqrt n})-P( z\le \dfrac{67-68}{2.5/\sqrt n})=0.95$

$\Rightarrow P(z\le \dfrac{\sqrt n}{2.5})-P( z\le \dfrac{-\sqrt n}{2.5})=0.95 \\ \Rightarrow 2P(z\le \dfrac{\sqrt n}{2.5})-1=0.95 \\ \Rightarrow P(z\le \dfrac{\sqrt n}{2.5})=0.975 \\ \Rightarrow P(z\ge \dfrac{\sqrt n}{2.5})=0.025 \\\Rightarrow \dfrac{\sqrt n}{2.5}=1.96 \\ \Rightarrow \sqrt n=4.9 \\ \Rightarrow n=4.9^2=24.01 \\ \Rightarrow n=25$