Question

Scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points. what is the minimum score you would need to be in the top 5%?

Answer 1

The top 5% of scores belong to the 95th percentile, which means the cutoff score $k$ is such that

$\mathbb P(X$

Transforming to the standard normal distribution, you have

$\mathbb P(X$

A left-tail probability of 95% corresponds to a z-score of about $k^*=1.6449$, which means the cutoff score must be around

$1.6449=\dfrac{k-76.4}{6.1}\implies k\approx86.4$