Answer to Question #172842 in Statistics and Probability for Kudzai

Question

If X is normally distributed with mean 150.3cm and variance 25. Find the probability that X is more than 10cm difference from the mean

Answer 1

$X\sim N(\mu, \sigma^2 )$

Given $\mu=150.3, \sigma^2=25$

$$P(X<140.3)=P(Z<\dfrac{140.3-150.3}{\sqrt{25}})$$

$$=P(Z<-2)\approx0.022750$$

$$P(X>160.3)=1-P(X\leq 160.3)$$

$$=1-P(Z\leq \dfrac{160.3-150.3}{\sqrt{25}})=1-P(Z\leq 2)$$

$$\approx0.022750$$

$$P(difference>10)$$

$$=P(X<140.30)+P(X>160.3)$$

$$\approx0.022750+0.022750=0.0455$$

Check

$\sigma=\sqrt{25}=5$

$10=2\sigma$

By the 68–95–99.7 rule

$$P(\mu-2\sigma\leq X\leq \mu+2\sigma)\approx0.9545$$

$$P(difference>10)\approx1-0.9545=0.0455$$