Section 9.5 The Normal Distribution

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1 Section 9.5 The Normal Distribution The normal distribution is a very common continuous probability distribution model in which most of the data points naturally fall near the mean, with fewer data points farther away from the mean. For example, if a test was given in a large class of students, most students would be expected to score at or near the mean, while fewer and fewer students would be expected to lie far from the mean. When graphed, a normal distribution takes on a recognizable bell curve shape: Figure: Typical normal distribution ( bell ) curve. Most of the data, represented as area, is clumped into the middle of the graph, while little of the data is found on the on the fringes. Since the curve is a probability density function, the area underneath is by definition equal to 1. The Standard Normal Distribution The standard normal distribution is a normal probability distribution function in which the mean is µ = 0, the variance is var = 1 and the standard deviation (the square root of the variance) is = 1. The notation for a normal distribution is N (µ, var), so the standard normal distribution is known as the N (0,1) distribution. The actual probability density function that defines the standard normal distribution is N(0,1) = f ( x) = 1 2π e 2 x, < x < This function is somewhat difficult to handle in an integral, so instead tables or calculators are used to determine the various areas underneath the probability density function. Since the standard normal distribution is a p.d.f., it has an area of 1 below its graph. The standard normal distribution N (0,1) is built in such a way that approximately 68% of the data is within one standard deviation of the mean, about 95% of the data is within two standard deviations of the mean, and over 99% of the data is within three standard deviations of the mean. Half of the area lies below (to the left) the mean, and half lies above (to the right) the mean. The middle hump of the distribution is sometimes called the body of the curve while the narrow fringes on both sides of the hump are called tails. The standard normal distribution has symmetry across the y-axis which can be used to help calculate areas over various bounds. 423

2 Figure: The shaded regions represent one, two and three standard deviations from the mean on the standard normal distribution. Example 1: Let r.v. X be normally distributed according to the N (0,1) distribution. Use the above graphs and symmetry to determine the following probabilities (areas): (a) P ( 1 X 0), (b) P ( 0 X 2), (c) P ( X 0) and (d) P ( X 1). Discussion: For (a), the area between 1 x 0 is half of the 0.68 that normally spans within one standard deviation of the mean. Hence, P ( 1 X 0) = For (b), the area over 0 x 2 is half of the 0.95 that normally spans within two standard deviations of the mean. Therefore, P ( 0 X 2) = Parts (c) and (d) are open-ended intervals. For part (c), P ( X 0) = 0. 5 since this encloses half of the full area under the graph. For (d), the area over 0 x 1 is 0.34, and added to the 0.5 over the left half of the graph, gives a total area of Therefore, P ( X 1) = The graphs of each probability from example 1 are on the next page. 424

3 Figure: Shaded areas for example 1. For standard deviation values that are not integers, a table or calculator is used to determine the areas over a given set of bounds, as the next example illustrates: Example 2: Use a table or calculator to determine these probabilities on the standard normal distribution N (0,1) : (a) P ( 0.45 X 1.26), (b) P ( 1 X 2.38) and (c) P ( X 0.35). Discussion: The TI models of calculator have a built-in feature under DISTR for calculating areas under a standard normal distribution: normalcdf(a,b). To calculate these probabilities as areas, simply call up the normalcdf command and type in the bounds, separated by a comma: (a) P ( 0.45 X 1.26) = normalcdf( 0.45, 1.26) = (b) P ( 1 X 2.38) = normalcdf( 1, 2.38) = (c) P ( X 0.35) = X 0.35) = 0.5 normalcdf(0, 0.35) = The calculator has the obvious advantage that the left and right bounds can be entered in together. Most tables are built to give readings from 0 to a positive x-value. Thus, to answer part (a) using a table, look up the area from 0 to 0.45, which by symmetry is the same as the area from 0.45 to 0. Then look up the area from 0 to 1.26, and finally sum the two values together. A similar approach will be needed if using a table for part (b). For (c), the interval is open-ended to the right, so we use the fact that the area to the right of the mean is 0.5. Look up the area from 0 to 0.35, and subtract from

4 Conversion to the Standard Normal Distribution: Z-Scores Most scenarios that follow a normal distribution do not have a convenient mean of µ = 0 and standard deviation of = 1. In order to use the table or calculators to solve general normal distribution scenarios, the input values x must be converted to fit onto the standard normal distribution curve. The results are called z-scores and are simply standard deviations on the standard normal distribution curve. The normalization formula is: z = x µ In practice this formula is very easy to use. Just be sure to convert any variance figures into standard deviation first. Example 3: Suppose r.v. X is normally distributed as a N (10,4) distribution. Find these probabilities: (a) P ( 9 X 11), (b) P ( 6.5 X 10.75) and (c), P ( X 9.8). Discussion: The mean of this distribution is µ = 10 and the standard deviation is = 4 = 2. To answer part (a), normalize the 9 and the 11 into their corresponding z-scores, and calculate: P ( 9 X 11) P Z = 0.5 Z 0.5) = For part (b), a similar conversion is made: P ( 6.5 X 10.75) P Z = 1.75 Z 0.375) = For part (c), the 9.8 is normalized into its z-score, and make note that the left endpoint is open ended: P ( X 9.8) P Z = Z 0.1) = Note that Z becomes a new random variable in place of X. The TI-83 calculator allows the mean and standard deviation to be entered in as parameters directly. For example, in the case P ( 9 X 11) above with mean µ = 10 and standard deviation = 2, the value can be determined by typing normalcdf(9,11,10,2), which gives the desired result of If the latter two parameters are not included, the calculator assumes the standard normal distribution of N (0,1). 426

5 Applications of the Normal Distribution The following examples illustrate the normal distribution as an application: Example 4: Suppose the average American male stands 70 inches tall (5 ft, 10 in) with a standard deviation of 2.5 inches. Assume the heights are normally distributed. A man is chosen at random. What is the probability he (a) stands between 5 ft, 6 in and 6 ft tall? (b) he stands at least 6 ft tall? And (c) he stands at most 5 ft, 9 in? Discussion: Let r.v. X represent the height of men in inches. For consistency, all heights are converted into inches, which in turn must be normalized into their corresponding z-scores. For (a), the probability is written P ( 66 X 72). With mean µ = 70 and = 2. 5, the heights are normalized into z-scores and the probability determined: P ( 66 X 72) P Z = 1.6 Z 0.8) = Therefore, about 73.3% of all American men stand between 5 ft 6 in and 6 ft tall. For (b), the probability is written P ( X 72), normalized into z-scores and the probability determined: P ( X 72) P Z = Z 0.8) = Therefore, about 21.2% of American men stand above 6 ft tall. For (c), the probability is written P ( X 69), normalized into z-scores and the probability determined: P ( X 69) P Z = Z 0.4) = About 34.5% of men stand at or below 5 ft, 9 in. In many applications, we may work the process in reverse using tables or the invnorm feature on the calculator. In cases like this, a percentage (area) underneath the normal distribution is known and we wish to work backwards to find the corresponding z-score that agrees with this area. Specifically, invnorm(a) returns a z-score on the standard normal curve N (0,1) such that the area from the left up to the z-score is the desired area a, which must be entered as a decimal, 0 < a < 1. invnorm(a) = z z N( 0,1) dx = a 427

6 Example 5: On the standard normal distribution N (0,1), what z-score corresponds to an area of 0.85, as read from the left? Discussion: This problem can be rephrased as P ( X z) = The invnorm feature is designed to return z-scores such that the area from the left up to the z-score is the desired percentage. Call up invnorm and enter in the area: invnorm(0.85) = Therefore, z = This z-score is graphed with its corresponding area: Remember, invnorm always returns z-scores for areas from the left up to the z-score. Example 6: On the standard normal distribution N (0,1), what z-score determines an area of 0.25 to the right of the z? In other words, what is the minimum z-score needed to be in the top 25%? Discussion: Graphically, the top 25% is the shaded area extending from (unknown) z and to the right: However, the InvNorm feature is designed to return areas from the left and up to z as a right endpoint. Therefore, the problem needs to be restated as invnorm(0.75), which is

7 Therefore, one needs to be positive standard deviation from the mean to be at the minimum threshold for the top 25%. Example 7: In a large college class, the mean score on the midterm was µ = 66 with a standard deviation of 8 points. The professor will give anyone who scored in the top 40% a B grade, and anyone who scored in the top 12% an A grade. What are the minimum scores needed for a B grade? an A grade? Discussion: This distribution is N (66,64). Be sure to note that var = 64, so that the standard deviation is = var = 64 = 8. In using the invnorm feature, we are assuming the N (0,1) standard normal distribution. Therefore, we will need to find the z-scores in N (0,1) then convert back to the N (66,64) setting. To be in the top 40% means you scored better than 60% of the rest of the students. On the standard normal distribution, InvNorm(0.6) = We need to convert this figure from the N (0,1) setting back to the N (66,64) by working the z-score formula in reverse: x µ z = = x 66 8 Solving for x, we get: x 66 = 0.253(8) = Therefore, the minimum score needed for a B on this test is a 68 (rounding as needed). For an A grade, the same approach is used: to be in the top 12% means the student is better than 88% of the students: invnorm(0.88) = Converting back to x, we get: x µ z = = x 66 8 Solving for x, we get x = Therefore, a minimum score of 75 (rounded) gives the student an A grade. Sometimes, grading on the curve means to adapt the scores onto a normal distribution such as the above example. The parameters for mean and standard deviation can also be included with the calculator command. For example, the above calculation could be entered as invnorm(0.88,66,8), which gives , or 75.4 as was determined. 429

8 Example 8: The average speed for vehicles along one notorious stretch of freeway is µ = 75 miles per hour with a standard deviation of = 6 miles per hour. The police will ticket anyone caught in the top 0.5% of speeders. What is the minimum speed one needs to go to get a ticket? Discussion: Being in the top 0.5% of speeders is equivalent to being faster than 99.5% of all drivers. Hence, invnorm(0.995) = This z-score is converted back to its corresponding x: x µ x 75 z = 2.58 = x = Therefore, a minimum speed of 90 miles per hour (rounded) earns you an automatic ticket. Example 9: You took a standardized exam and scored 95 points. The mean was µ = 71. Your friend scored 80 points and was found to be in the 85 th percentile (meaning he scored better than 85% of all other test takers). What is your percentile? Discussion: The only missing piece of information here is the standard deviation. Your friend s result can be used to determine the standard deviation. Since he was in the 85 th percentile, your friend s z-score is invnorm(0.85) = Working this into the z-score conversion formula with unknown, we can solve for : x µ z = = = 9 = = Therefore we can determine your z-score: z = = Thus, your percentile is equivalent to the area below z = : Therefore, you scored in the 96.7 th percentile. p (Z 1.842) = Two final comments regarding the Normal distribution: P ( X = k) = 0 for all k, for the reasons outlined in section 9.4. The inequalities and <, and and >, are used interchangeably: X < k) = X k), and so forth. Inclusion or exclusion of the endpoint has no effect on the probability. By default, the inequalities and are used. Thus, a statement such as X is at most 20 can be interpreted as P ( X 20) or P ( X < 20) ; both are the same. 430

9 Summary The standard normal distribution is a bell-shaped curve with mean µ = 0, variance var = 1 and standard deviation = 1. Any normal distribution is written N (µ, var). The standard normal distribution is called the N (0,1) distribution. The standard normal distribution is a probability density function defined for < x < and an area of 1 below the graph. Probabilities are read as areas under the curve. About 68% of the data is within one standard deviation of the mean, 95% of the data is within two standard deviations of the mean, and over 99% of the data is within three standard deviations of the mean. To 3-decomal place accuracy: 1 X 2 X 3 X 1) = ) = ) = The standard normal distribution is symmetric: the area to the left of the mean µ = 0 is 0.5 and to the right is 0.5. X X 0) = 0.5 0) = 0.5 To convert from a general normal distribution with mean µ and standard deviation to corresponding z-scores on the N (0,1) distribution, use the normalization formula z = x µ invnorm(a) returns a z-score on the standard normal curve N (0,1) such that the area from the left up to the z-score is the desired area a, which must be entered as a decimal, 0 < a < 1. invnorm(a) = z z N( 0,1) dx = a 431