If the height of $300$ students are normally distributed with mean $64.5$ inches and standard deviation $3.3$ inches, find the height below which $99\%$ of the students lie.
Mean height of students $= \mu = 64.5$ inches
Standard deviation of the height of students $= \sigma = 3.3$ inches
Let $X$ be a normal variate.
Standard normal variate $Z = \dfrac{X - \mu}{\sigma}$ $\;\;\; \cdots \; (1)$
$99\%$ of students corresponds to an area of $0.99$ under the normal probability curve.
Area to the left of $Z = 0$ is $= 0.5$
Remaining area $= 0.99 - 0.5 = 0.49$
Area of $0.49$ corresponds to $Z = 2.33$ $\;\;\;$ [from $z$ tables]
From equation $(1)$, $X = \left(Z \times \sigma\right) + \mu$
$\therefore \;$ When $Z = 2.33$, $\;$ $X = \left(2.33 \times 3.3\right) + 64.5 = 72.189$
$\therefore \;$ Height below which $99\%$ of the students lie $= 72.189 \approx 72.19$ inches