Solutions to Empirical Rule Practice Problems

1. What percent of a normal population is more than 2 standard deviations above the mean?
P( z > 2 ) =

Solution: Since the problem asks about an abstract “normal population”, you have only z scores and no x scores (real-world data).

Begin with your sketch, as shown at right. First, mark z = 2, which is the dividing line for 2 standard deviations above the mean. Since you want more than 2 standard deviations above the mean, shade the right-hand tail.

How to find this area? You know the 68–95–99.7 rule (Empirical Rule), which tells you that 95% of the area of the curve or 95% of the data will lie between 2 standard deviations above and below the mean. So you draw an auxiliary line at z = −2 and mark the area of 95% in the center.

But if the center area is 95%, then the area of the two tails combined is 100%−95% = 5%. And since the curve is symmetric, the two tails must have equal area, so the area of one tail is 5%÷2 = 2.5%.

Answer: P(z > 2) = 2.5%

Note: All of the above text is just commentary to help you understand the solution process. The diagram itself is perfectly adequate to show the work and the answer for this problem.

2. What percent of adult women have heights between 65.5 and 68 in (5′5½″ and 5′8″)?
P( 65.5 ≤ x ≤ 68 ) =

Solution: 65.5 in is the mean (z = 0), and 68 in is one standard deviation above the mean (z = (68−65.5)/2.5 = 1). Sketch the curve, showing both the x axis and the z axis.

You need an auxiliary line at z = −1. Then the center area is 68%. The area you’re actually interested in is half that: 68%÷2 = 34%.

Answer: 34%

3. What proportion of adult women have heights greater than 70.5 in (5′10½″)?
P( x > 70.5 ) =

Answer: 2.5% or 0.025

4. What’s the probability that a randomly selected adult woman is less than 63 in tall (5′3″)?
P( x < 63 ) =

Answer: 16% or 0.16

5. If a woman is 68 in (5′8″) tall, what’s her percentile rank?
P( x _____ 68 ) =

Solution: First, you have to remember that “percentile” means the percent of the sample or population that has a lower score. So you are looking for P(x < 68).

Now you can draw your sketch. z = (68−65)/2.5 = 1. Since you need x < 68, you shade the area to the left of 68 in. (This is one of those cases where you end up shading more than half of the curve.)

The boundary is z = 1, so draw an auxiliary line at z = −1. Now you have two areas: the left half of the curve, which is 50%, and the part from z = 0 to z = 1. That part must be half of the part from z = −1 to z = 1: 68%÷2 = 34%. Adding, you have 34%+50% = 84%.

Answer: She is at the 84th percentile, or P₈₄.

6. What’s the area under the curve between 58 in and 68 in (4′10″ and 5′8″)?
P( 58 < x < 68 ) =

Solution: The z scores are −3 and +1, so you want P(−3 < z < 1). Make your sketch, and draw auxiliary lines at −1 and +3.

There are several possible ways to proceed, but I think the simplest is to divide the shaded area at z = 0.

You know that the area under the curve from z = −3 to z = +3 is 99.7%. Half of that, 49.85%, is the area from z = −3 to z = 0.

You also know that the area from z = −1 to z = +1 is 68%. The area from z = 0 to z = 1 is half that, 68%÷2 = 34%.

Adding 49.85%+34%, you have the Answer: 83.85% or 0.8385.

Interpretation: 83.85% of women are between 58 in and 68 in tall, or if you randomly select one woman, there is an 83.85% chance that she is between 58 in and 68 in tall.

7. What proportion of adult women are between 5′5½″ and 5′6½″?
P( 65.5 ≤ x ≤ 66.5 ) =

Answer: 66.5″ corresponds to z = 0.4. At this time you have enough information to solve only problems with z scores of whole numbers −3 to +3, so this problem has no solution at present.

See also: Later, you’ll learn how to do Normal Calculations on TI-83/84 or TI-89.