# If $X$ a normal variate with mean $80$ and standard deviation $10$ compute the following probabilities by standardizing.$P(85\leq$X$\leq$95$)$

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}}$$e^{-\Large\frac{1}{2}z^2};-\infty < Z < \infty • Z\sim N(0,1) Step 1: X\sim N(80,10^2) Let Z=\large\frac{X-\mu}{\sigma} The probability density function Z is \phi(z)=\large\frac{1}{\sqrt{2\pi}}$$e^{-\Large\frac{1}{2}z^2};-\infty < Z < \infty$
Step 2:
$P(85 \leq X\leq 95)$
When $X=85$
$Z=\large\frac{85-80}{10}$$=0.5 When X=95 Z=\large\frac{95-80}{10}$$=1.5$
$P(85 \leq X \leq 95)=P(0.5 \leq Z \leq 1.5)$
$\qquad\qquad\qquad\;\;=P(0\leq Z\leq 1.5)-P(0\leq Z \leq 0.5)$
$\qquad\qquad\qquad\;\;=0.4332-0.1915$
$\qquad\qquad\qquad\;\;=0.2417$

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edited Oct 31, 2013 by pady_1