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

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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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answered Sep 19, 2013 by sreemathi.v
edited Oct 31, 2013 by pady_1
 

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