Question

Suppose that the mean and standard deviation of the scores on a statistics exam are 78 and 6.11, respectively, and are approximately normally distributed. Calculate the proportion of scores above 74.

Answer 1

Answer:

0.7422 = 74.22% of scores are above 74.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 78, \sigma = 6.11$

Calculate the proportion of scores above 74.

This is 1 subtracted by the pvalue of Z when X = 74. So

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

$Z = \frac{74 - 78}{6.11}$

$Z = -0.65$

$Z = -0.65$ has a pvalue of 0.2578

So 1-0.2578 = 0.7422 = 74.22% of scores are above 74.