Question

If the test scores in Personal Finance Unit 9 are normally distributed with a

mean of 87 and a standard deviation of 5, what percentage of the scores
were between 82 and 92?

Answer 1

68.269%

Explanation:

We start by calculating the z-scores of both numbers

Mathematically;

z-score = (x-mean)/SD

where x represents the raw score

For 82;

z-score = (82-87)/5 = -5/5 = -1

For 92;

z-score = (92-87)/5 = 5/5 = 1

So the probability we want to calculate would be;

P(-1 < x< 1)

We can use the standard normal distribution table to get this.

By using the distribution, P(-1<x<1) = 0.68269

By converting this to percentage, we have 68.269%