Question

A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that a randomly selected score will be greater than 63.7

Answer 1

Answer:

93.32% probability that a randomly selected score will be greater than 63.7.

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:

$\mu = 80.2, \sigma = 11$

What is the probability that a randomly selected score will be greater than 63.7.

This is 1 subtracted by the pvalue of Z when X = 63.7. So

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

$Z = \frac{63.7 - 80.2}{11}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

1 - 0.0668 = 0.9332

93.32% probability that a randomly selected score will be greater than 63.7.