# If $X$ a normal variate with mean $80$ and standard deviation $10$ compute the following probabilities by standardizing. $P(65\leq$X$\leq$100$)$

## 1 Answer

Toolbox:
• Standard normal distribution:
• In a standard normal distribution $\mu=0,\sigma ^2=1$
• The random variable $X$ can be converted to the standard normal variable $Z$ by the transformation
• $Z=\large\frac{X-\mu}{\sigma}$
• The probability density function $Z$ is $\phi(z)=\large\frac{1}{\sqrt{2\pi}}$$e^{-\Large\frac{1}{2}z^2};-\infty < Z < \infty • Z\sim N(0,1) Step 1: X\sim N(80,10^2) Let Z=\large\frac{X-\mu}{\sigma} \qquad=\large\frac{X-80}{10} be the standard normal variate. Step 2: P(65 \leq X\leq 100) When X=65 Z=\large\frac{65-80}{10}$$=-1.5$
When $X=100,Z=2$
$\therefore P(65 \leq X \leq 100)=P(-1.5\leq Z \leq 2)$
$\qquad\qquad\qquad\quad\;\;=P(-1.5\leq Z\leq 0)+P(0\leq Z\leq 2)$
$\qquad\qquad\qquad\quad\;\;=P(0\leq Z\leq 1.5)+P(0\leq Z\leq 2)$
$\qquad\qquad\qquad\quad\;\;=0.4332+0.4772$
$\qquad\qquad\qquad\quad\;\;=0.9104$
answered Sep 19, 2013

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