(i) The mean and mode both equal the median; that is, the average value and the most likely value are both in the middle of the distribution.

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1 MATH 382 Normal Distributions Dr. Neal, WKU Measurements that are normally distributed can be described in terms of their mean µ and standard deviation σ. These measurements should have the following properties: (i) The mean and mode both equal the median; that is, the average value and the most likely value are both in the middle of the distribution. (ii) The measurements are symmetric about the mean. (iii) (The Rule): Around 68% of the measurements should be within one standard deviation of average, around 95% should be within two standard deviations of average, and around 99.7% of the measurements should be within three standard deviations of average. (iv) A histogram of measurements create a Bell-Shaped Curve with the percentages at the high and low ends dropping off exponentially. Such a measurement is denoted by X ~ N (µ, σ ). When µ = 0 and σ = 1, then we have the standard normal distribution that is denoted by Z ~ N (0, 1). Typical Histogram Theoretical Shape Example 1. In the U.S., birth weights are normally distributed with a mean of 7 pounds and a standard deviation of 2 pounds. Explain what this means in terms of the properties of a normal distribution. Solution. Let Ω be all babies born in the U.S., and let X denote the birth weight. Then X ~ N (7, 2) lbs. That is, (i) In the U.S., the average birth weight, the most likely birth weight, and the median birth weight are all 7 lbs. (ii) Birth weights as a whole are symmetric about the weight 7 lbs. (iii) Around 68% of newborn weights are from 5 to 9 lbs (µ ± σ ); around 95% of newborn weights are from 3 to 11 lbs (µ ± 2σ ); and around 99.7% of newborn weights are from 1 to 13 lbs (µ ± 3σ ). (iv) A histogram of newborn birth weights create a Bell-Shaped Curve with the percentages of high and low weights dropping off exponentially.

2 The theoretical bell-shaped curve is defined by the graph of the probability density function 1 f X (x ) = σ 2π e (x µ )2 / 2σ 2, for all x, which is symmetric about the mean µ. The probability that a measurement falls between values a and b is given by the area under this graph from point a to point b and is denoted by ( ) = f X ( x) dx P a X b b = a 1 σ 2π b e ( x µ )2 / 2σ 2 dx. a In addition, tail-value probabilities are of the form P(X k) = k f X ( x) dx and P(X k) = f X (x ) dx. P( a X b) P(X k) P(X k) k a b µ µ k k Normal Calculations Unfortunately, the above integrals have no closed-form antiderivatives and therefore cannot be evaluated by hand. But given a normal distribution X ~ N (µ, σ ), we do wish to calculate these types of probabilities. These probabilities can be computed with the built-in normalcdf( command from the DISTR menu on a TI84: P(a X b) = normalcdf(a, b, µ, σ ) P(X < k) = P(X k) = normalcdf( 1 E99, k, µ, σ ) P(X > k) = P(X k) = normalcdf(k, 1 E99, µ, σ ) The bounds 1 E99 and 1 E99 are used as estimates of and +. Inverse Normal Calculation To find the value x for which P(X x ) equals a desired probability p (an inverse normal calculation), we use the command invnorm( p,µ,σ ). The invnorm( command is also found in the DISTR menu. Note: In quantile notation, x = π p.

3 Example 2. The lengths of human pregnancies are approximately normally distributed with a mean of 266 days and a standard deviation of 16 days. (a) What is the population Ω? What is the measurement X and its distribution? (b) What percent of pregnancies last at most 240 days? (c) What percent of pregnancies last from 240 to 270 days? (d) How long do the longest 20% of pregnancies last? Solution. (a) Here Ω is the population of all women who have given birth and X is the measurement of how many days the pregnancy lasted. Then X N(266, 16). (b) For X ~ N(266, 16), we wish to find P(X 240). We use the command normalcdf( 1E99, 240, 266, 16) to obtain P(X 240) So only about 5.2% of pregnancies last at most 240 days. (c) To find P(240 X 270), enter the command normalcdf(240, 270, 266, 16). We see that around 54.66% of pregnancies last from 240 to 270 days (d) To find how long the longest 20% of pregnancies last, we must find the value x for which P(X x ) = But to use the invnorm( command, we instead must find x such that P(X x ) = Using the command invnorm(.80, 266, 16), we see that the longest 20% of pregnancies last at least days. Example 3. Heights of adult women are normally distributed with a mean of 65.5 inches and a standard deviation of 2.75 inches. What percentage of women are (a) at least 70 in. tall? (b) at most 63 in. tall? (c) from 64 to 68 in. tall? (d) What height is such that 95% of all women are below this height? (e) What height is such that 90% of all women are above this height? (f) What two heights, symmetric about the mean, contain 50% of all heights? Solution. Here, Ω = All adult women and X = height in inches. Then X ~ N(65.5, 2.75). (a) At least 70, meaning 70 or more: Enter the command normalcdf(70, 1 E99, 65.5, 2.75) to obtain P(X 70) So about 5.09% of women are at least 70 inches tall.

4 (b) At most 63, meaning up to 63: Enter the command normalcdf( 1 E99, 63, 65.5, 2.75) to obtain P(X 63) So about % of women are at most 63 inches tall. ( c) P(64 X 68) So about 52.56% of women are from 64 to 68 inches tall. (d) We must find x such that P(X < x) = To do so, enter invnorm(.95, 65.5, 2.75) to obtain the height of x 70 in. for which 95% of women measure below. (e) We must find x such that P(X > x) = 0.90 or equivalently such that P(X x) = To do so, enter invnorm(.10, 65.5, 2.75) to obtain the height of x in. for which 90% of women measure above. (f) If we want 50% of heights in the middle, then we need x and y such that P(x X y) = But then we need 25% of the heights at each tail. So we need x such that P(X x) = 0.25 and y such that P(X y) = Enter invnorm(.25, 65.5, 2.75) to obtain x inches and invnorm(.75, 65.5, 2.75) to obtain y in. The Standard Normal Distribution Suppose X ~ N (µ, σ ) is a normally distributed measurement. For example IQ scores are such that X ~ N (100, 15). Then most measurements (about 99.7%) are within µ ± 3σ, which is 55 to 145 for IQ scores. If we subtract µ from every measurement, then we still have a normal distribution, but most values will be between 3σ and 3σ. By subtracting µ, the result is N (0, σ ). Next, suppose we divide the new values by σ. Then most values will be between 3 and 3. The result is now N (0, 1). By subtracting µ from every measurement and then dividing by σ, we have standardized the values and have obtained the standard normal distribution Z ~ N (0, 1), also called the Gaussian distribution. Let X ~ N (µ, σ ) and Z = X µ σ. Then Z ~ N (0, 1) and f Z ( x) = 1 2π e x2 / 2.

5 For Z ~ N (0, 1), we still can compute the various probabilities and inverse calculations using the normalcdf( and invnorm commands. For example, P(Z 1.22) and P( 0.85 Z 1.05) are shown below. 1 1 Z ~ N (0, 1) P(Z 1.22) P( 0.85 Z 1.05) Example 4. Let Z ~ N (0, 1). (a) Find the number z such that P(Z z) = (b) Find the numbers w and z such that P(w Z z) = Solution. For Part (a), we actually need P(Z z) = So enter the command invnorm(.95, 0, 1) to obtain z For Part (b), we need w and z such that P(Z w) = and P(Z z) = So enter the commands invnorm(.025, 0, 1) and invnorm(.025, 0,1 ) to obtain w 1.96 and z Note that w = z by the symmetry about 0 of the N (0, 1) curve. By converting different normal distributions X and Y (having different units) to standard normal distributions, we place X and Y on the same scale (each with no units). Values from X and Y then can be compared without any probability calculation. Example 5. Adult female heights are X ~ N(66, 2.4) inches and baby birth weights are Y ~ N(7, 2) lbs. Which value is more extreme: An adult female height of 72 in. or a new-born weight of 4 pounds? Solution. Simply convert each value to a standard normal scale: X = 72 in Z = 72 in 66 in 2.4 in = 2.5 and Y = 4 lb Z = 4 lb 7 lb 2 lb = 1.5 Because the standardized value of 2.5 is further from 0 on a standard scale than 1.5 is, we see that the adult female height of 72 in. is the more extreme measurement.

6 Exercises 1. Students in a Psychology Masters Program are given an IQ test. The scores are generally found to be normally distributed with a mean of 112 and a standard deviation of 9. (a) Give the population Ω under consideration and the measurement X. What is the notation for the distribution of X? Give the pdf of X. (b) Compute (i) P(X 105) (ii) P(100 X 124) (iii) P(X 130) (c) What score do only 5% of these students score at least as high as? (d) What scores x and y, symmetric about the mean, are such that P(x X y) = 0.66? 2. Let Z ~ N (0, 1). (a) Compute (i) P(Z 1. 45) (ii) P(Z 2.12) (iii) P( 2 Z 2) (b) Find the numbers w and z such that P(w Z z) = (c) Find the number z such that P(Z z) = ACT scores are X ~ N(22.4, 3.2) and SAT scores are Y ~ N(1020, 160). (a) Which is a better score, an ACT of 28 or an SAT of 1400? (b) Which happens less often, an ACT of at most 14 or an SAT of at least 1400? 1 4. Let f (x) = σ 2π e (x µ )2 / 2σ 2, for all x, be the pdf of the N(µ, σ ) distribution. Derive the values of x that give the inflection points of f ( x).

7 Addendum: z scores Let Z ~ N(0, 1) be the standard normal distribution having mean 0 and standard deviation 1. The z-score for probability r is the value z such that P ( z Z z) = r. That is, there is probability r between the points z and + z. Prob = r z z We denote the total remaining outer probability by the symbol α, where r = 1 α. Then there is α /2 probability at each tail. In this case, the z-scores are usually denoted by z α /2, where α /2 refers to the right-tail probability. Here are the most commonly used z-scores: r = 1 α α/2 z α /2 z α /2 Inner Prob = 1 α Tail Prob = α/2 z α / Note: Any other z-score can be found with the built-in invnorm( command from the DISTR menu. For example, if we want the 80% z-score, then there is 80% probability in the middle region. So there is α /2 = 10% at each tail. The (positive) z-score then comes from finding the inverse of 90% cumulative probability of the standard normal distribution and is approximately z 0.10 = α/2 10% + 80% invnorm(.90, 0, 1) 1.28 Helpful Fact: For a general normal distribution, X ~ N(µ, σ ), the bounds that contain inner probability r = 1 α are given by µ ± z α/ 2 σ.

8 Example. Heights of adult women are normally distributed with a mean of 65.5 inches and a standard deviation of about 3 inches. Use z-scores to solve the following: (a) What heights are such that 90% of all women are between these heights? (b) What heights are such that 95% of all women are between these heights? (c) What heights are such that 75% of all women are between these heights? Solution. (a) The bounds that contain 90% of (normally distributed) measurements are µ ±1.645σ, which here are 65.5 ± Thus, 90% of women are from to inches in height. (b) The bounds that contain 95% of the measurements are µ ±1.96σ, which here are 65.5 ± Thus, 95% of women are from to inches in height. (c) First, we need the 75% z-score. If there is 75% inner probability, then there is 12.5% at each tail. Hence, the (positive) z-score comes from finding the inverse of 87.5% cumulative probability of the standard normal distribution and is about So the bounds that contain 75% of the measurements are µ ± 1.15σ, which here are 65.5 ± Thus, 75% of women are from to inches in height. Exercise IQ scores are generally found to be normally distributed with a mean of 100 and a standard deviation of 15. Use z -scores to find the bounds that contain (a) 95% of all scores (c) 40% of all scores (b) 98% of all scores (b) 94% of all scores