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

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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}}$$e^{-\Large\frac{1}{2}z^2};-\infty < Z < \infty$
  • $Z\sim N(0,1)$
Step 1:
$X\sim N(80,10^2)$
Let $Z=\large\frac{X-\mu}{\sigma}$
$\qquad=\large\frac{X-80}{10}$ be the standard normal variate.
Step 2:
$P(X\leq 100)$ When $X=100,Z=\large\frac{100-80}{10}$$=2$
$\therefore P(X\leq 100)=P(Z\leq 2)=0.5$+area beneath the curve between $Z=0$ and $Z=2$
$\Rightarrow 0.5+0.4772=0.9772$
answered Sep 19, 2013 by sreemathi.v
 

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