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 percent of students scored below 95 points?

Answer 1

Answer:

99.89% of students scored below 95 points.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 76.4, \sigma = 6.1$

What percent of students scored below 95 points?

This is the pvalue of Z when X = 95. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{95 - 76.4}{6.1}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989.

99.89% of students scored below 95 points.