Students pass a test if they score 50% or more. The marks of a large number of students were sampled and the mean and standard deviation were calculated as 42% and 8% respectively. Assuming this data is normally distributed, what percentage of student pass the test?

let X be the sore random variable.

$\begin{aligned} &E(X)=42 \%=0.42 \\ &S D(X)=8 \%=0.08 \\ &P(X>0.50)=? \end{aligned}$

We know that if $X \sim N(\mu, \sigma)$ then,

$\begin{aligned} z&=\frac{X-\mu}{\sigma} \sim N(0,1) \\ &\text{So,}\ P(\frac{X-0.42}{0.08}>\frac{0.50-0.42}{0.08})\\ &=P(z>1) \end{aligned}$

From z-table calculator.

$P(X>0.50)=P(z>1)=0.15866$

So $15.866 \%$ students will pass the test.