155S6.3_3 Applications of Normal Distributions
MAT 155 Statistical Analysis
Dr. Claude Moore
Cape Fear Community College
Chapter 6
Normal Probability Distributions
6­1 Review and Preview
6­2 The Standard Normal Distribution
6­3 Applications of Normal Distributions
6­4 Sampling Distributions and Estimators
6­5 The Central Limit Theorem
6­6 Normal as Approximation to Binomial
6­7 Assessing Normality
September 30, 2011
Technology for this chapter in order of priority (easy of use) are
1. Excel programs (created by Dr. Moore)
2. TI calculator
3. Statdisk (CD in textbook or download through CourseCompass
4. GeoGebra program
See Excel programs of Normal Distribution (xls) under the Technology on the Important Links webpage.
See TI tutorials of normcdf and invNorm under the TI Calculator on the Important Links webpage.
Use Statdisk that came with your textbook or download Statdisk from CourseCompass.
Statdisk is very similar to Table A­2 in terms of what is need in order to use it.
http://cfcc.edu/mathlab/geogebra/normal_curve_aba.html
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155S6.3_3 Applications of Normal Distributions
Key Concept
September 30, 2011
Conversion Formula
This section presents methods for working with normal distributions that are not standard. That is, the mean is not 0 or the standard deviation is not 1, or both.
The key concept is that we can use a simple conversion that allows us to standardize any normal distribution so that the same methods of the previous section can be used.
Round z scores to 2 decimal places
Converting to a Standard Normal Distribution
Example – Weights of Water Taxi Passengers In the Chapter Problem, we noted that the safe load for a water taxi was found to be 3500 pounds. We also noted that the mean weight of a passenger was assumed to be 140 pounds. Assume the worst case that all passengers are men. Assume also that the weights of the men are normally distributed with a mean of 172 pounds and standard deviation of 29 pounds. If one man is randomly selected, what is the probability he weighs less than 174 pounds?
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155S6.3_3 Applications of Normal Distributions
Example ­ cont September 30, 2011
Example ­ cont
P ( x < 174 lb.) = P(z < 0.07)
= 0.5279
Helpful Hints
1. Don’t confuse z scores and areas. z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A­2 lists z scores in the left column and across the top row, but areas are found in the body of the table.
2. Choose the correct (right/left) side of the graph.
Procedure for Finding Values Using Table A­2 and Formula 6­2
1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the x value
(s) being sought.
2. Use Table A­2 to find the z score corresponding to the cumulative left area bounded by x. Refer to the body of Table A­2 to find the closest area, then identify the corresponding z score. 3. A z score must be negative whenever it is located in the left half of the normal distribution. 3. Using Formula 6­2, enter the values for µ, s, and the z score found in Step 2, then solve for x.
x = µ + (z • s) (Another form of Formula 6­2)
(If z is located to the left of the mean, be sure that it is a negative number.)
4. Areas (or probabilities) are positive or zero values, but they are never negative. 4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem.
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155S6.3_3 Applications of Normal Distributions
September 30, 2011
Example – Lightest and Heaviest
Example – Lightest and Heaviest ­ cont
Use the data from the previous example to determine what weight separates the lightest 99.5% from the heaviest 0.5%? x = µ + (z ● σ)
x = 172 + (2.575 29)
x = 246.675 (247 rounded)
TI: x = invNorm(area to left, mean, std)
TI: x = invNorm(area to left, mean, std)
Example – Lightest and Heaviest ­ cont
The weight of 247 pounds separates the lightest 99.5% from the heaviest 0.5%
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155S6.3_3 Applications of Normal Distributions
278/6. IQ Scores. In Exercises 5–8, find the area of the shaded region. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Use Table A­2 & TI.
September 30, 2011
278/8. IQ Scores. In Exercises 5–8, find the area of the shaded region. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Use Table A­2 & TI.
TI: P(x > 80) = normalcdf(small, large, mean, std)
TI: P(x > 80) = normalcdf(small, large, mean, std)
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155S6.3_3 Applications of Normal Distributions
278/10. IQ Scores. In Exercises 9­12, find the indicated IQ score. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Use TI calculator.
September 30, 2011
278/12. IQ Scores. In Exercises 9­12, find the indicated IQ score. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Use TI calculator.
TI: x = invNorm(area to left, mean, std)
TI: x = invNorm(area to left, mean, std)
IQ Scores. In Exercises 13–20, assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). (Hint: Draw a graph in each case.)
278/14. Find the probability that a randomly selected adult has an IQ greater than 131.5 (the requirement for membership in the Mensa organization).
Use TI calculator.
IQ Scores. In Exercises 13–20, assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). (Hint: Draw a graph in each case.)
278/16. Find the probability that a randomly selected adult has an IQ between 110 and 120 (referred to as bright normal). Use TI calculator.
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155S6.3_3 Applications of Normal Distributions
IQ Scores. In Exercises 13–20, assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). (Hint: Draw a graph in each case.)
279/18. Find the first quartile Q1, which is the IQ score separating the bottom 25% from the top 75%. Use TI calculator.
In Exercises 21–26, use this information (based on data from the National Health Survey): • Men’s heights are normally distributed with mean 69.0 in. and standard deviation 2.8 in. • Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in.
279/22. Doorway Height The Gulfstream 100 is an executive jet that seats six, and it has a door­way height of 51.6 in. a. What percentage of adult men can fit through the door without bending? b. What percentage of adult women can fit through the door without bending? c. Does the door design with a height of 51.6 in. appear to be adequate? Why didn’t the engineers design a larger door? d. What doorway height would allow 60% of men to fit without bending?
September 30, 2011
IQ Scores. In Exercises 13–20, assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). (Hint: Draw a graph in each case.)
279/20. Find the IQ score separating the top 37% from the others.
Use TI calculator.
In Exercises 21–26, use this information (based on data from the National Health Survey): • Men’s heights are normally distributed with mean 69.0 in. and standard deviation 2.8 in. • Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in.
279/24. Tall Clubs International Tall Clubs International has minimum height requirements for men and women. a. If the requirements are changed so that the tallest 4% of men are eligible, what is the new minimum height for men? b. If the requirements are changed so that the tallest 4% of women are eligible, what is the new minimum height for women?
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155S6.3_3 Applications of Normal Distributions
In Exercises 21–26, use this information (based on data from the National Health Survey): • Men’s heights are normally distributed with mean 69.0 in. and standard deviation 2.8 in. • Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in.
279/25. U. S. Army Height Requirements for Women The U. S. Army requires women’s heights to be between 58 in. and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join the Army because they are too short or too tall? b. If the U. S. Army changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?
280/30. Aircraft Seat Width Engineers want to design seats in commercial aircraft so that they are wide enough to fit 99% of all males. (Accommodating 100% of males would require very wide seats that would be much too expensive.) Men have hip breadths that are normally distributed with a mean of 14.4 in. and a standard deviation of 1.0 in. (based on anthropometric survey data from Gordon, Clauser, et al.). Find P99. That is, find the hip breadth for men that separates the smallest 99% from the largest 1%.
September 30, 2011
In Exercises 21–26, use this information (based on data from the National Health Survey): • Men’s heights are normally distributed with mean 69.0 in. and standard deviation 2.8 in. • Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in.
279/26. Marine Corps Height Requirement for Men The U. S. Marine Corps requires that men have heights between 64 in. and 80 in. a. Find the percentage of men who meet the height requirements. Are many men denied the opportunity to become a Marine because they do not satisfy the height requirements? b. If the height requirements are changed so that all men are eligible except the shortest 3% and the tallest 4%, what are the new height requirements?
280/32. Sitting Distance A common design requirement is that an item (such as an aircraft or theater seat) must fit the range of people who fall between the 5th percentile for women and the 95th percentile for men. If this requirement is adopted, what is the minimum sitting distance and what is the maximum sitting distance? For the sitting distance, use the buttock­to­knee length. Men have buttock­to­knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. Women have buttock­to­knee lengths that are normally distributed with a mean of 22.7 in. and a standard deviation of 1.0 in.
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