Question

The scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. What percent of the scores are less than 81?The scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. What percent of the scores are less than 81?

Answer 1

We are given the following variables:

Normal distribution

u = sample mean = 74

s = standard deviation of the samples = 7

x = sample value = 81

Since we are asked to find percent of the scores less than 81, this means we have to find the proportion using the z-score. The formula for z score is:

z = (x – u) / s

Substituting the given values:

z = (81 – 74) / 7

z = 1

Using the standard normal distribution tables for z, we find that:

P = 0.8413            at z = 1

Therefore this means that 84.13% of scores are less than 81.

Answer 2

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