If $X$ is a normal variate with mean $80$ and standard deviation $10$, compute the following probabilities by standardizing:
- $P \left(X \leq 100\right)$
- $P \left(85 \leq X \leq 95\right)$
Given: Normal variate $= X$; mean $= \mu = 80$; standard deviation $= \sigma = 10$
Standard normal variate $Z = \dfrac{X - \mu}{\sigma}$
- When $X = 100$, $\;$ $Z = \dfrac{100 - 80}{10} = 2$
$\begin{aligned} \therefore \; P \left(X \leq 100\right) & = P \left(Z \leq 2\right) \\\\ & = \left(\text{area between } Z = - \infty \text{ to } Z = 0\right) \\\\ & \hspace{1.5cm} + \left(\text{area between } Z = 0 \text{ to } Z = 2\right) \\\\ & = 0.5000 + 0.4772 \;\;\; \left[\text{from Normal Distribution Table}\right] \\\\ & = 0.9772 \end{aligned}$
- When $X = 85$, $\;$ $Z = \dfrac{85 - 80}{10} = 0.5$
When $X = 95$, $\;$ $Z = \dfrac{95 - 80}{10} = 1.5$
$\begin{aligned} \therefore \; P \left(85 \leq X \leq 95\right) & = P \left(0.5 \leq Z \leq 1.5\right) \\\\ & = \left(\text{area between } Z = 0 \text{ to } Z = 1.5\right) \\\\ & \hspace{1.5cm} - \left(\text{area between } Z = 0 \text{ to } Z = 0.5\right) \\\\ & = 0.4332 - 0.1915 \;\;\; \left[\text{from Normal Distribution Table}\right] \\\\ & = 0.2417 \end{aligned}$