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 =

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) 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) 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)

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).

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).

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?

Example 4: Use the following information to answer the questions that follow: *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. The Mark VI monorail used at 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?

Example 4: Use the following information to answer the questions that follow: *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. The Mark VI monorail used at 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?

Helpful Hints 1. Don’t confuse z scores/x - values with 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, µ, σ) 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. A= 0.5675 What IQ score is this? Since we are looking for a score, we use the invNorm feature (2nd Vars). Type in: invNorm(amount of area on the left of the missing score, mean, standard deviation) It will look like this: invNorm(.5675, 100, 15) ANSWER: 102.6

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. A= 0.90 A= 0.10 What IQ score is this? Since we are looking for a score, we use the invNorm feature (2nd Vars). Type in: invNorm(amount of area on the left of the missing score, mean, standard deviation) It will look like this: invNorm(.9, 100, 15) ANSWER: 119.2

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. 69 A= 0.95 Start by drawing a bell curve, labeling the mean: The 95th percentile means that 95% of heights are lower than the missing height. Mark where the height will be that separates the lower 95% of heights with the upper 5% of heights. How much area is on the left of the missing height? Since we are looking for a height, we use the invNorm feature (2nd Vars). Type in: invNorm(amount of area on the left of the missing height, mean, standard deviation) It will look like this: invNorm(.95, 69, 2.8) ANSWER: Doorways should be 73.6 inches high to allow 95% of men to fit without bending or bumping their head.

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. 3420 A= 0.99 Start by drawing a bell curve, labeling the mean: A= 0.01 A= 0.03 We are looking for 2 different weights. Mark where the weight will be that separates the lower 3% of weights and another mark where the weight will separate the upper 1% of weights. How much area is on the left of the first missing weight? How much area is on the left of the second missing weight? (1 – 0.01) The first weight’s set up will look like this: invNorm(.03, 3420, 495) The second weight’s set up will look like this: invNorm(.99,3420,495) ANSWER: Birth weights less than 2489 g and higher than 4571.5 g should receive special treatment.

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%. b) Find the IQ score separating the top 28% from the others. 100 A= 0.72 Start by drawing a bell curve, labeling the mean: A= 0.43 The 43rd percentile means that 43% of scores are lower than the missing score. Mark where the score will be that separates the lower 43% of scores with the upper 57% of heights. A= 0.28 How much area is on the left of the missing height? Your set up will look like this: invNorm(.43, 100, 15) ANSWER: P43 = 97.4 The top 28% means that 72% of scores are lower than the missing score. Mark where the score will be that separates the lower 72% of scores with the upper 28% of heights. How much area is on the left of the missing height? Your set up will look like this: invNorm(.72, 100, 15) ANSWER: 108.7

Example 10: Use the following information to answer the questions that follow: *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. Start by drawing a bell curve, labeling the mean and the 2 heights that make up the height requirement: 63.6 80 58 We are looking for a percentage (or area), which is the normalcdf set up from yesterday. Women that meet the height requirement are between 58 and 80 inches. Type in: normalcdf (left score, right score, mean, standard deviation) Your set up will look like this: normalcdf(58,80,63.6,2.5) ANSWER: 0.9875, which is 98.75%

Example 10: Use the following information to answer the questions that follow: *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. 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? 63.6 A= 0.98 Start by drawing a bell curve, labeling the mean: A= 0.02 Now we are looking for 2 different heights. Mark where the height will be that separates the lower 1% of heights and another mark where the height will separate the upper 2% of heights. A= 0.01 How much area is on the left of the first missing height? How much area is on the left of the second missing height? (1 – 0.02) The first height’s set up will look like this: invNorm(.01, 63.6, 2.5) The second height’s set up will look like this: invNorm(.98, 63.6, 2.5) ANSWER: The new height requirement would allow women between 57.8 and 68.7 inches into the U.S. Army