Applications of Normal Distributions Section 6-3 Applications of Normal Distributions

Key Concept 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.

Conversion Formula Round z scores to two decimal places.

Converting to a Standard Normal Distribution x –   z =

When finding areas with a nonstandard normal distribution, use this procedure: 1. Sketch a normal curve, label the mean and specific x values, then shade the region representing the desired probability. 2. Use the normalcdf feature on your calculator to find the area of the shaded region. This area is the desired probability.

P(x < a) : denotes the probability that the z score is less than a. Method for Finding Nonstandard Normal Distribution Areas or Probabilities (When being asked to find an area or probability you will use the same method) In the following notations, x represents a non-standardized individual value. P(a < x < b) : denotes the probability that the z score is between a and b. To find this probability in your calculator, type: normalcdf(a, b, µ, σ) a   b P(x > a) : denotes the probability that the z score is greater than a. To find this probability in your calculator, type: normalcdf(a, 99999, µ, σ) a   P(x < a) : denotes the probability that the z score is less than a. To find this probability in your calculator, type: normalcdf(–99999, a, µ, σ) a  

Example 1: Find the area of the shaded region Example 1: Find the area of the shaded region. The graph depicts 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). P(x < 110) = Normalcdf(–99999, 110, 100, 15) = 0.7475

Example 2: Find the area of the shaded region Example 2: Find the area of the shaded region. The graph depicts 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). P(85 < x < 125) = Normalcdf(85, 125, 100, 15) = 0.7936

Example 3: The typical home doorway has a height of 6 ft. 8 in Example 3: The typical home doorway has a height of 6 ft. 8 in., or 80 in. Because men tend to be taller than women, we will consider only men as we investigate the limitations of the standard doorway height. Given the heights of men are normally distributed with a mean of 69.0 in. and a standard deviation of 2.8 in., find the percentage of men who can fit through the standard doorway without bending or bumping their head. Is that percentage high enough to continue using 80 in. as the standard height? P(x < 80) = Normalcdf(–99999, 80, 69, 2.8) = 0.99996 Yes, almost all men will be able to fit through the door, only less than 1% will have to bend down to fit through the door.

Example 4: Use the following information the answer the questions that follow: *Men’s heights are normally distributed with mean 69.0 in. and standard deviation 2.8 in. The Mark VI monorail used a Disney World and the Boeing 757-200ER airliner have doors with a height of 72 in. a) What percentage of adult men can fit through the doors without bending? P(x < 72) = Normalcdf(–99999, 72, 69, 2.8) = 0.8580

Example 4 cont.: Use the following information the answer the questions that follow: *Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in. The Mark VI monorail used a Disney World and the Boeing 757-200ER airliner have doors with a height of 72 in. b) What percentage of adult women can fit through the doors without bending? P(x < 72) = Normalcdf(–99999, 72, 63.6, 2.5) = 0.9996

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. 2. Choose the correct (right/left) side of the graph. A value separating the top 10% from the others will be located on the right side of the graph, but a value separating the bottom 10% will be located on the left side of the graph. 3. A z score must be negative whenever it is located in the left half of the normal distribution. 4. Areas (or probabilities) are positive or zero values, but they are never negative.

Method for Finding Values from a Nonstandard Normal Distribution when given an Area or Probability 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. Determine the area to the left and enter the following into your calculator: InvNorm(area to the left, mean, standard deviation) 3. 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. page 263 of Elementary Statistics, 10th Edition

Example 5: Find the indicated IQ score Example 5: Find the indicated IQ score. The graph depicts 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). The shaded area is 0.5675. invNorm(0.5675, 100, 15) = 102.55

Example 6: Find the indicated IQ score Example 6: Find the indicated IQ score. The graph depicts 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). The shaded area is 0.10. The shaded area is to right so you must find the area to the left: 1 – 0.10 = 0.90, now you can enter the following command into your calculator: invNorm(0.90, 100, 15) = 119.22

Example 7: When designing an environment, one common criterion is to use a design that accommodates 95% of the population. How high should doorways be if 95% of men will fit through without bending or bumping their head? That is, find the 95th percentile of heights of men. Heights of men are normally distributed with a mean of 69.0 in. and a standard deviation of 2.8 in. Percentiles are area to the left so you can enter the following command into your calculator: invNorm(0.95, 69, 2.8) = 73.61 in.

Example 8: The Newport General Hospital wants to redefine the minimum and maximum birth weights that require special treatment because they are unusually low or unusually high. After considering relevant factors, a committee recommends special treatment for birth weights in the lowest 3% and the highest 1%. The committee members soon realize that specific birth weights need to be identified. Help this committee by finding the birth weights that separate the lowest 3% and the highest 1%. Birth weights in the United States are normally distributed with a mean of 3420 g and a standard deviation of 495 g. The lowest 3% is going to be area to the left, so: invNorm(0.03, 3420, 495) = 2489 g. The highest 1% is the area to the right, so area to the left will be 1 – 0.01 = 0.99, so: invNorm(0.99, 3420, 495) = 4571.54 g.

Example 9: 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). a) Find P43 which is the IQ score separating the bottom 43% from the top 57%. Remember that percentiles are area to the left, so: invNorm(0.43, 100, 15) = 97.35 b) Find the IQ score separating the top 28% from the others. The top 28% is the area to right so the area to the left is 72%, so: invNorm(0.72, 100, 15) = 108.74

Example 10: Women’s heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in. 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. Normalcdf(58, 80, 63.6, 2.5) = 0.9875 b) If the U.S. Army changes the height requirements so that all women are eligible except the shortest 1% and tallest 2%, what are the new height requirements? The shortest 1% is going to be area to the left, so: invNorm(0.01, 63.6, 2.5) = 57.78 in. The tallest 2% is the area to the right, so area to the left will be 1 – 0.02 = 0.98, so: invNorm(0.98, 63.6, 2.5) = 68.73 in.