**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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