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7.3: Inverse Normal Calculations

Created by: Bruce DeWItt
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Learning Objectives

  • Understand how to use the Normal Distribution Table and the z-score formula to find values for a particular normal distribution given a percentile
  • Be able to use the Inverse Normal command on a graphing calculator to find values for a particular normal distribution given a percentile
  • Be able to find values for a particular normal distribution given a 'middle' percentage range

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.

License: CC BY-NC 3.0
[Figure2]


Begin by finding the value closest to .8000 in the Normal Distribution Table. We find our closest value to be .7995 which corresponds to a z-score of 0.84. Put this value into the z-score formula to get 0.84=\frac{x-10}{4}.

0.84=\frac{x-10}{4}

3.36=x-10

13.36=x

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.

License: CC BY-NC 3.0

It is easy to calculate that marker line 'a' is at the 20th percentile and marker line 'b' is at the 80th percentile simply by noting their relationship to the 50th percentile marker. In addition, note that 'a' and 'b' clearly enclose the middle 60 percent of all data. From the Normal Distributions Table, we can see that the z-score associated with the 20th percentile is -0.84 and the z-score associated with the 80th percentile is 0.84. We now calculate -0.84=\frac{x-335}{14} or x = 303.2 yards. A similar calculation at the 80th percentile gives us x = 326.8 yards. 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 \mu = 196 kg and \sigma = 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.

0.84=\frac{x-196}{11}

9.24=x-196

205.24=x

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

Problem Set 7.3

Exercises

1) The Standard Normal Curve is defined as having a mean of 0 and a standard deviation of 1.

a) What is the z-score associated with a result at the 84th percentile?

b) What is the z-score associated with a result at the 16th percentile?

c) Find a z-score such that only 5% of the Standard Normal Curve is to the right of that z-score.

d) Find a z-score such that only 35% of the Standard Normal Curve is the left of that z-score.

e) Find the two z-scores such that the middle 50% of the Standard Normal Curve is between the two z-scores.

2) Doctors often monitor their patients blood-glucose levels. It is known that for blood-glucose levels, \mu = 85 and \sigma = 25.

License: CC BY-NC 3.0

a) Draw and label sketch of the normal distribution for this situation marking the mean and 1, 2, and 3 standard deviations above and below the mean. [Figure5]

b) It turns out that doctors consider the blood-glucose level of a patient to be normal if the level is in the middle 94% of all results. What range of blood-glucose levels constitute the middle 94% of all results?

c) Patients are considered to be high risk for diabetes if their blood-glucose test comes back in the top 1% of all results. What blood-glucose level marks the start of the top 1% of blood-glucose levels?

d) Doctors also show concern if there is too little blood-glucose in a patient's system. They will prescribe treatments to patients if their blood-glucose is in the lowest 2% of all patients. What is the blood-glucose level that marks this boundary?


3) For a given population of high school juniors and seniors, the SAT math scores are normally distributed with a mean of 500 and a standard deviation of 100. For that same population, the ACT math exam has a mean of 18 with a standard deviation of 6.

a) One school requires that students score in the top 10% on their SAT math exam for admission. What is the minimum score that a student must achieve to be considered for this school?

b) Another school requires that students score in the top 40% on their ACT math exam for admission. What is the minimum score that a student must achieve to be considered for this school?

c) One particular school likes to focus on mid-level students and so they only accept students who are in the middle 50% of all ACT math test takers. Between what two scores must a student achieve in order to be considered for acceptance into this school?

d) One student boasts that they scored at the 85th percentile on their ACT math exam. Another student brags that they scored a 620 on the SAT math exam. Who did better?

4) Many athletes train to try to be selected for the US Olympic team. Suppose for the men's 100 meters, the athletes being considered for the team have a mean time of 10.06 seconds with a standard deviation of 0.07 seconds. In the final qualifying event for the team, only the top 20% of runners will be selected. What time must a runner get to be in the top 20%?

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5) A high school basketball coach notices that taller players tend to have more success on his team. As a result, the coach decides that only the tallest 25% of the boys in the 11th and 12th grades will be allowed to try out for the team this year. Suppose that the mean height of 11th and 12th grade boys is 5 feet 9 inches with a standard deviation of 2.5 inches. How tall must a player be in order to be able to try out for the team?

6) A student comes home to his parents and excitedly claims that he is in the top 90% of his class. Explain why this might not be worth getting excited about.

7) At a certain fast-food restaurant, automatic soft drink filling machines have been installed. For 20-ounce cups, the machine is set to fill up the cups with 19 ounces of soda. Unfortunately, the machine is not perfectly consistent and does not always dispense 19 ounces of soda. Suppose the amount it dispenses produces a normal distribution with a mean of 19 ounces and a standard deviation of 0.6 ounces. It turns out that the 20 ounce cup will actually hold a bit more than 20 ounces. A mathematically inclined worker notices this and starts to record what happens when the machine fills the cups. It turns out that the cups overfill 2% of the time. How much soda will the 20-ounce cup actually hold?


Review Exercises

8) Adult male American bald eagles have a mean wingspan of 79 inches with a standard deviation of 3.5 inches. What percent of these eagles have wingspans longer than 7 feet?

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9) Consider the data in the table below where the number of pages is the explanatory variable.

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a) Create a scatterplot for the data set. Label your axes. [Figure8]

b) Determine the correlation coefficient, r, for the scatterplot.

c) Give the least-squares linear regression equation. Be sure to define your variables.

d) Using your answer from part c), predict the weight of a book that has 130 pages.

e) Using your answer from part c), predict the number of pages for a book that weighs 295 grams.

10) Consider a standard set of 15 pool balls. Pool balls #1-#8 are solid and pool balls #9-#15 are striped.

a) If you randomly select one pool ball, what is the probability that it is both solid and odd?

b) If you randomly select one pool ball, what is the probability that it is either solid or odd?

c) If you randomly select two pool balls without replacement, what is the probability that they are either both solid or both striped?


Image Attributions

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Date Created:

Jun 14, 2011

Last Modified:

Oct 01, 2014
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