C.6 Normal Distributions APPENDIX C.6 Normal Distributions A43 Find probabilities for continuous random variables. Find probabilities using the normal distribution. Find probabilities using the standard normal distribution. Find probabilities using -scores for nonstandard normal distributions. Continuous Random Variables As discussed in Appendi C.5, a random variable associated with an eperiment that has a finite number of possible outcomes is called a discrete random variable. For eample, if a coin is tossed three times and a random variable is assigned that counts the number of heads that turn up, then there are only four values the random variable can have 0,, 2, or 3. (see Figure C.25.) 0 2 3 FIGURE C.25 In this appendi you will study eperiments for which there are an infinite number of possible outcomes. For eample, consider an eperiment that measures the actual number of ounces of toothpaste in a tube that is labeled 0 ounces. The actual weight could be 0.02 ounces, or 0.024 ounces, or 0.0247 ounces, and so on. There are infinitely many different possible weights. Suppose that the possible weights vary between 9.5 ounces and 0.5 ounces. The range of the random variable needed for this eperiment would be 9.5 0.5. Range of random variable This type of random variable is a continuous random variable because it can have any value in the interval between 9.5 ounces and 0.5 ounces. (See Figure C.26.) 9.5 0 0.5 FIGURE C.26 Because there are an infinite number of values in the interval between 9.5 and 0.5, you cannot list each possible value of the random variable and assign probabilities to those values (as you did for a discrete random variable). To measure probabilities for a continuous random variable, you must use a method that involves measuring the area of a region under a graph.
A44 APPENDIX C Probability and Probability Distributions A function that is used to measure probabilities for a continuous random variable is called a probability density function. Three common types of probability density functions are uniform, eponential, and normal probability density functions, as shown in Figure C.27. f() b a Area = a Uniform Probability Density Function b f() Area = Eponential Probability Density Function f() Area = Normal Probability Density Function FIGURE C.27 For each of the probability density functions shown in Figure C.27, the range of the continuous random variable is an interval on the real line. For the uniform probability density function, the range is the closed interval a, b, which implies that the random variable can take on any value between a and b (including a or b). For the eponential probability density function, the range is the interval 0,, which implies that the random variable can take on any nonnegative value. For the normal probability density function, the range is the interval,, which implies that the random variable can take on any value. As indicated in Figure C.27, the area of the region that lies between the graph of a probability density function and the -ais is. This corresponds to the fact that, for a discrete random variable, the sum of all probabilities must be.
( APPENDIX C.6 Normal Distributions A45 Probability 60 f() FIGURE C.28 Uniform Probability Density Function Width = 0 Height = 60 50 60 70 80 90 200 20 Flight time (in minutes) Eample Uniform Probability Density Function The flight time for a trip from Atlanta to Pittsburgh on a certain airline has been found to vary between 2.5 hours (50 minutes) and 3.5 hours (20 minutes). Moreover, the probability of the flight time falling in any given one-minute interval in this range is the same. That is, the model used to represent the probabilities is the uniform probability density function f 60, 50 20. (See Figure C.28.) Find the probability that the flight time for a flight is between 70 minutes and 80 minutes. SOLUTION From Figure C.28, note that the area of the region under the graph of the uniform probability density function is. You can verify this by using the formula for the area of a rectangle: P 50 20 width height 20 50 60. To find the probability that the flight time is between 70 and 80 minutes, find the area of a smaller rectangle whose width is the interval 70, 80. That is, P 70 80 width height 80 70 60 0.67. CHECKPOINT In Eample, what is the probability that a flight will take more than 3 hours? All uniform probability density functions have the form f b a, a b. Uniform probability density function As shown in Figure C.29, the width of the rectangular region corresponding to a uniform probability density function is b a and the height is b a. So, the area is P a b b a. b a Uniform Probability Density Function f() b a P(c d) = (d c) ( b a a c d b FIGURE C.29
A46 APPENDIX C Probability and Probability Distributions So, for a uniform probability density function, the probability that lies in the interval c d (where a c d b) is P c d d c b a. The epected value (or mean) of a uniform probability density function is a b 2, the midpoint of the interval a, b. Try showing that P a a b 2 2 and P a b 2 b 2. From this discussion about uniform probability density functions, you should observe the following important facts, which apply to all probability density functions.. With a probability density function, the probability that the random variable will have an eact value is insignificant. So, the probability that the value of the random variable lies in a particular interval is found. 2. The area of the region under the graph of a probability density function is. 3. The probability that a continuous random variable will lie in an interval is defined to be the area of the region between the interval on the -ais and the graph of the probability density function. f() μ 3σ μ 2σ μ σ μ + μ + 2σ μ σ μ+ 3σ Approimately 68% of data items Approimately 95% of data items Nearly all data items (99.74%) FIGURE C.30 34.% 3.6% 2.2% Normal Distribution In practice, the most widely used probability density function is the normal probability density function. This function can be applied to many practical problems that occur in the life and social sciences. The equation that is used to represent a normal probability density function is f 2 e 2 2 2 Normal probability density function You will see, however, that it is usually not necessary to use this equation to solve problems involving normal distributions. The graph of a normal probability density function is called the normal curve, and the associated probability distribution is called the normal distribution. A normal curve is bell-shaped, as shown in Figure C.30. Note that the curve is symmetric about the mean. The percentages shown in this figure are valid for all normal distributions. For eample, for all normal distributions, the probability that an outcome will fall between and is 0.34. Also, approimately 95% of the total region under a normal curve lies within two standard deviations of the mean. STUDY TIP For a more detailed discussion of epected value and standard deviation of a continuous random variable, see Section.3.
APPENDIX C.6 Normal Distributions A47 Eample 2 Normal Distribution Probability Tire Lifetime f() 34.% 34.% 3.6% 3.6% 2.2% 2.2% 42,500 47,500 52,500 57,500 Tire lifetime (in miles) FIGURE C.3 The lifetimes of tires are normally distributed with a mean of 50,000 miles and a standard deviation of 2500 miles. Use Figure C.30 to answer the following questions. a. What is the probability that a tire will last for 50,000 miles or more? b. What is the probability that a tire will last for 52,500 miles or less? c. What is the probability that a tire will last between 47,500 miles and 55,000 miles? SOLUTION Using the probabilities shown in Figure C.30, you can obtain the graph and probabilities shown in Figure C.3. a. Because the distribution is normal, half of the area under the curve lies to the right of the mean. So, the probability that a tire will last for 50,000 or more miles is P 50,000 0.500. 2 b. Because half of the tires will last for 50,000 miles or less, and 34.% will last between 50,000 miles and 52,500 miles, the probability that a tire will last for 52,500 miles or less is approimately P 52,500 0.500 0.34 0.84 or approimately 84.%. c. Using Figure C.3, the probability that a tire will last between 47,500 miles and 55,000 miles is approimately P 47,500 55,000 0.34 0.34 0.36 0.88 or approimately 8.8%. CHECKPOINT 2 A medical researcher has determined that for a group of 00 women, the lengths of pregnancies from conception to birth are normally distributed with a mean of 266 days and a standard deviation of 2 days. a. What is the probability that a pregnancy will last for 266 days or more? b. What is the probability that a pregnancy will last for 278 days or less? c. What is the probability that a pregnancy will last between 266 and 278 days?
A48 APPENDIX C Probability and Probability Distributions The Standard Normal Distribution A continuous random variable that has a normal distribution for which and is said to have a standard normal distribution, and the associated probability density function is called the standard normal probability density function. For this function, it is customary to represent the random variable by the letter (called a -score). Note in Figure C.32 that the units of the horiontal ais represent the number of standard deviations from the mean. 0 f() = e 2 /2 2 π f() 0.4 0.2 Area = μ = 0 σ = 0. 3 2 2 3 Standard Normal Distribution FIGURE C.32 To find probabilities involving the standard normal distribution, it is convenient to use a table such as the one shown in Figure C.34 on the net page. This table has 30 entries which correspond to -scores from 0.00 to 3.09 in increments of 0.0. Note in this table that the tenths-digit is read down the left column and the hundredths-digit is read across the top row. For eample, to find the probability for a -score of 2.4, think of 2.4 as 2. 0.04; then find the value that is in the row labeled 2. and the column labeled 0.04. In this case, that value is 0.4838. This result means that the probability that the random variable has a -score between 0 and 2.4 is 0.4838. That is, P 0 2.4 0.4838 as shown in Figure C.33. 0.4 P(0 2.4) = 0.4838 f() 0.2 0. 3 2 2 3 2.4 FIGURE C.33
APPENDIX C.6 Normal Distributions A49 FIGURE C.34 Standard Normal Table (-Scores) f() -score Hundredths Digit for -score.00.0.02.03.04.05.06.07.08.09 0.0.0000.0040.0080.020.060.099.0239.0279.039.0359 0..0398.0438.0478.057.0557.0596.0636.0675.074.0753 0.2.0793.0832.087.090.0948.0987.026.064.03.4 0.3.79.27.255.293.33.368.406.443.480.57 0.4.554.59.628.664.700.736.772.808.844.879 0.5.95.950.985.209.2054.2088.223.257.290.2224 0.6.2257.229.2324.2357.2389.2422.2454.2486.257.2549 0.7.2580.26.2642.2673.2704.2734.2764.2794.2823.2852 0.8.288.290.2939.2967.2995.3023.305.3078.306.333 0.9.359.386.322.3238.3264.3289.335.3340.3365.3389.0.343.3438.346.3485.3508.353.3554.3577.3599.362 Tenths Digit for -score..3643.3665.3686.3708.3729.3749.3770.3790.380.3830.2.3849.3869.3888.3907.3925.3944.3962.3980.3997.405.3.4032.4049.4066.4082.4099.45.43.447.462.477.4.492.4207.4222.4236.425.4265.4279.4292.4306.439.5.4332.4345.4357.4370.4382.4394.4406.448.4429.444.6.4452.4463.4474.4484.4495.4505.455.4525.4535.4545.7.4554.4564.4573.4582.459.4599.4608.466.4625.4633.8.464.4649.4656.4664.467.4678.4686.4693.4699.4706.9.473.479.4726.4732.4738.4744.4750.4756.476.4767 2.0.4772.4778.4783.4788.4793.4798.4803.4808.482.487 2..482.4826.4830.4834.4838.4842.4846.4850.4854.4857 2.2.486.4864.4868.487.4875.4878.488.4884.4887.4890 2.3.4893.4896.4898.490.4904.4906.4909.49.493.496 2.4.498.4920.4922.4925.4927.4929.493.4932.4934.4936 2.5.4938.4940.494.4943.4945.4946.4948.4949.495.4952 2.6.4953.4955.4956.4957.4959.4960.496.4962.4963.4964 2.7.4965.4966.4967.4968.4969.4970.497.4972.4973.4974 2.8.4974.4975.4976.4977.4977.4978.4979.4979.4980.498 2.9.498.4982.4982.4983.4984.4984.4985.4985.4986.4986 3.0.4987.4987.4987.4988.4988.4989.4989.4989.4990.4990
A50 APPENDIX C Probability and Probability Distributions P(0.02) 3 2 2 3 (a) P(.43 0) 3 2 2 3 (b) P(0.4.23) 3 2 2 3 (c) P( 0.76) Eample 3 Using the Standard Normal Tables Use Figure C.34 to determine the following probabilities. a. P 0.02 b. P.43 0 c. P 0.4.23 d. P 0.76 SOLUTION a. To determine P 0.02, find the value that is in the row labeled.0 and the column labeled.02. That value is 0.346. So, P 0.02 0.346 b. To determine P.43 0, first use the symmetry of the standard normal probability density function to conclude that P.43 0 P 0.43. Then, using the row labeled.4 and the column labeled.03, you can conclude that P.43 0 P 0.43 0.4236. c. To determine P 0.4.23, use the fact that P 0.4.23 P 0.23 P 0 0.4. From Figure C.34, P 0.23 0.3907 and P 0 0.4 0.0557. So, P 0.4.23 0.3907 0.0557 0.3350. d. To determine P 0.76, use the fact that P 0 P 0 0.76 P 0.76. Because P 0 0.5000 and P 0 0.76 0.2764, you can conclude that P 0.76 0.5000 0.2764 0.2236. The areas corresponding to these four probabilities are shown in Figure C.35. 3 2 2 3 (d) FIGURE C.35 CHECKPOINT 3 Use Figure C.34 to determine the following probabilities. a. P 0 0.98 b. P 2.4 0 c. P 0.55.06 d. P 0.37
APPENDIX C.6 Normal Distributions A5 -Scores for Nonstandard Normal Distributions To convert measurments from a nonstandard normal distribution to the standard normal distribution, you must convert from -values to -scores, using the formula -score. Note that a normal distribution is considered to be nonstandard if its mean is not ero or if its standard deviation is not. For eample, suppose a nonstandard normal distribution has a mean of and a standard deviation of To find the probability that a random variable will have -values in the interval 5, 6.7, determine the -scores for 5 and 6.7: 5 and So, nonstandard -score for 5 5 5 2 -score for 6.7 Using Figure C.34, as shown in Figure C.36. 0. 6.7 5 2 0.85. standard 2. P 5 6.7 P 0 0.85. P 5 6.7 P 0 0.85 0.3023 f() μ = 5 σ = 2 Nonstandard μ = 0 σ = f() Standard P(5 6.7) = P(0 0.85) 2 3 4 5 6 7 2 2 = 6.7 = 0.85 FIGURE C.36
A52 APPENDIX C Probability and Probability Distributions Eample 4 Finding Probabilities for Nonstandard Distributions The mean speed of traffic (as checked by radar) on a stretch of interstate highway is 68 miles per hour with a standard deviation of 4 miles per hour. Assuming that the speeds of vehicles on this section of highway are normally distributed, determine what percentage of the vehicles are eceeding the legal speed of 65 miles per hour. SOLUTION Because the mean is 68 and the standard deviation is 4, the -score for 65 miles per hour is 65 68 4 0.75. Using Figure C.34, the percentage of vehicles is approimately P > 65 P 0.75 P 0.75 0 P 0 P 0 0.75 P 0 0.2734 0.5000 0.7734 77.34%. CHECKPOINT 4 In Eample 4, what percentage of vehicles are traveling between 65 and 70 miles per hour? CONCEPT CHECK. A random variable can take on any value in a given interval. 2. What is the total area of the region under the graph of a probability density function? 3. In the normal probability density function f e 2 2 2, which variable represents the mean? Which variable represents the standard deviation? 4. A continuous random variable that has a normal distribution for which and is said to have the normal distribution. 0 2
APPENDIX C.6 Normal Distributions A53 Skills Review C.6 The following warm-up eercises involve skills that were covered in earlier sections. You will use these skills in the eercise set for this section. For additional help, review Sections. and.2. In Eercises and 2, use the discrete probability distribution to determine the indicated probabilities.. 0 2 3 4 P 2 6 3 6 6 6 3 6 2 6 (a) P 2 (b) P > 2 2. 2 0 2 P 0.078 0.234 0.376 0.234 0.078 (a) P (b) P 0 In Eercises 3 6, determine whether the function f represents a probability density function over the given interval. If f is not a probability density function, identify the conditions(s) that is (are) not satisfied. 3. f 9,, 0 4. f, 0, 0 5. f 3 2e 3 2, 0, 6. f 0.006 25 2, 5, 5 Eercises C.6 In Eercises 4, find the indicated probabilities for the given uniform probability density function.. f 45, 45 90 (a) P 70 (b) P 50 60 (c) P 50 75 (d) P 80 2. f 90, 30 220 (a) P 30 (b) P 220 (c) P 68 20 (d) P 30 220 3. f 30, 95 25 (a) P 2 20 (b) P 0 25 (c) P 0 30 (d) P 95 4. f 25, 75 00 (a) P 00 (b) P 00 (c) P 75 (d) P 75 5. Departure Time Buses arrive and depart from a college every 20 minutes. The probability density function for the waiting times t (in minutes) for people arriving at random is f t 20, 0 t 20. See www.calcchat.com for worked-out solutions to odd-numbered eercises. Find the probability that a person will wait (a) no more than five minutes, and (b) at least 2 minutes. 6. Flight Time The flight times for trips from Miami to Atlanta on an airline has been found to vary between 90 minutes and 60 minutes. The probability density function for the flight times t (in minutes) is f t 70, 90 60. Find the probability that a flight will take (a) between 00 and 0 minutes, and (b) more than 2 hours. In Eercises 7 6, use the standard normal table in Figure C.34 to find the indicated probability. 7. P 0 0.93 8. P 0 2.05 9. P.0 0 0. P 0.34 0.96. P.5 0.07 2. P 2.42.2 3. P.93 4. P 2.7 5. P 0.52 6. P.5
A54 APPENDIX C Probability and Probability Distributions In Eercises 7 24, find the indicated probability for a continuous random variable that is normally distributed with a mean of 30 and standard deviation of 5. 7. P 30 35 8. P 30 37 9. P 23 30 20. P 8 30 2. P 20 40 22. P 25 35 23. P 9 36 24. P 2 45 In Eercises 25 32, find the indicated probability for a continuous random variable that is normally distributed with a mean of 5 and a standard deviation of 3. 25. P 8 3 26. P 7 27. P 7 20 28. P 9 22 29. P 8 30. P 20 3. P 9 32. P 9 33. Monthly Utility Bill The monthly utility bills in a city are normally distributed with a mean of $60 and a standard deviation of $6. What is the probability that the utility bill for a household in the city, chosen at random, is (a) between $56 and $74, (b) less than $54, and (c) between $65 and $70? 34. Subscription Rates The weekly health magaine sales per sales-person for a publishing company are normally distributed with a mean of 50 subscriptions and a standard deviation of 4 subscriptions. In any one week, what is the probability that a salesperson, chosen at random, will sell (a) between 39 and 45 subscriptions, (b) more than 47 subscriptions, and (c) more than 56 subscriptions? 35. Quality Control The lengths of hypodermic needles produced by a medical supply manufacturer are normally distributed with a mean length of inch and a standard deviation of 0.02 inch. What is the probability that a needle will have a length greater than.02 inches? 36. Quality Control One-liter bottles on an assembly line are filled by a liquid-dispensing machine. The amounts of liquid in the bottles are normally distributed with a mean of liter and a standard deviation of 0. liter. What is the probability that the machine fills a bottle with less than 0.95 liter? 37. Income Variation The annual incomes of associate professors at a college are normally distributed with a mean of $66,300 and a standard deviation of $5000. What is the probability that an associate professor at the college has a salary higher than $70,000? 38. SAT Scores SAT math scores for a group of high school seniors are normally distributed with a mean of 500 and a standard deviation of 00. What percentage of the students have scores (a) below 400, (b) above 700, and (c) between 500 and 600? In Eercises 39 44, find the indicated probabilities using the eponential probability density function. f t e t, Remember from Section.2 that the probability that t lies in the interval [c, d] is d P c } t } d f t dt. c 3, P t 3. 3 P 0 t., 5 2, 39. If find 40. If find 4. If find P 0.5 t 2. 42. If find P 0. t 0.2. 0.06, [0,. 43. Waiting Time The waiting times t (in minutes) for patients arriving at a doctor s office are eponentially distributed with Find the probability of waiting (a) less than 0 minutes, and (b) between 0 and 20 minutes. 44. Shelf Life The shelf lifes (in years) of a drug are eponentially distributed with A pharmacy has a quantity of this drug in stock and plans to replace the supply during regularly scheduled inventory replenishment periods. How much time should elapse between inventory replenishment periods if at least 95% of the supply is to remain within the shelf life throughout the period? 6. 4.5. In Eercises 45 50, use a symbolic integration utility to find the indicated probability for a random variable that has the standard normal distribution. Compare your results to those given in the standard normal table (Figure C.34). 45. P 0.0 2.0 46. P 2.0 47. P.5 2.0 48. P.5 0.3 49. P 0.3 2.0 50. P.0 5. Quality Control The lengths of fleible plastic catheters produced by a medical supply manufacturer are normally distributed with a mean length of 4 inches and a standard deviation of 0.07 inch. Use a symbolic integration utility to determine the probability that a catheter will have a length (a) less than 3.9 inches, (b) greater than 4. inches, and (c) between 3.9 and 4. inches. 52. SAT Scores Nationally in 2006, the SAT math scores roughly followed a normal distribution with a mean of 58 and a standard deviation of 5. Using a symbolic integration utility, find the percentage of students who had scores (a) below 400, (b) above 700, and (c) between 500 and 600. Compare your results to those of Eercise 38. Would you say that the students scores in Eercise 38 were better or worse than the 2006 national scores? Eplain. (Source: College Board)