7.3 Notes/Examples

We can now comfortably calculate percentages, percentiles, and probabilities given key information about a normal distribution. It is possible to go the other direction. In other words, if you are told a certain result is at a specific percentile, you can figure out what the actual value is equal to that is at that percentile. The process can be done using the Normal Distribution Table in Appendix A, Part 2. Begin by identifying the percentile you are interested in and finding it in the table. From there, put the value from the table into the z-score formula and solve it for the observation in question.

Example 1

Suppose that 10th grade girls have hair lengths that are normally distributed with a mean of 10 inches and a standard deviation of 4 inches. How long would a 10th grade girl's hair have to be in order to be at the 80th percentile for length?

Solution

The figure below shows the distribution of hair lengths and also marks where the 80th percentile is located.

A 10th grade girl would have to have a hair length of about 13.4 inches to be at the 80th percentile. This looks to be right based upon comparison to the figure above.

Technology

Once again, it is important to note that technology can be used to solve these types of problems without having to reference the Normal Distribution Table. The command that is commonly used for these types of problems is the Inverse Normal command or InvNorm. The Inverse Normal command requires users to enter the percentile in question, the mean, and the standard deviation. To solve the problem in Example 1, we could have typed in InvNorm(0.80,10,4) and we would have immediately had an answer of 13.366 or about 13.4 inches of hair.

Be sure you know how to access this command if you have a graphing calculator. Appendix C has some notes for users of graphing calculators.An online calculator that can produce the same information can be found at http://wolframalpha.com .

'Middle' and 'Top' Problems

Finally, we are sometimes in situations where we want to know what range of results are found in a middle percentage interval or what value one would have to be at in order to be in a specific top percentage. For example, a car salesman might wish to know what sales prices comprise the middle 50% of his sales to help him learn more about who his customers tend to be or a student might wish to know what they need to score on a test in order to be in the top 10%. Once again, this process can be done with either the Normal Distributions Table or by using technology.

Example 2

Professional golfer John Daly is known for his long drives off the tee. Suppose his drives have a mean distance of 315 yards with a standard deviation of 12 yards. What lengths of drives will constitute the middle 60% of all of his drives?

Solution

The sketch below is helpful in understanding what is happening here.

In other words, the middle 60% of John Daly's drives will travel between 303.2 yards and 326.8 yards.

We also could have used the Inverse Normal command once we knew the percentiles. InvNorm(.20,335,14) = 323.2 yards and InvNorm(.80,335,14) = 346.8 yards.

Example 3

In a weightlifting competition, the amount that the competitors can lift is normally distributed with a mean of 196 kg and a standard deviation of 11 kg.

Only the top 20% of all competitors will be able to advance to the next phase of the competition. What amount must a competitor lift in order to move into the next phase of the competition?

Solution

The key to this problem is noticing that to be in the top 20%, a competitor would actually have to be at the 80th percentile. The z-score at the 80th percentile is z=0.84.

The competitor would have to lift about 205 or 206 kg. Using a calculator, we get InvNorm(.8,196,11) = 205.26 kg.

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