1. 8  Distributions of Random Variables  Expected Value  Variance and Standard Deviation  The Binomial Distribution  The Normal Distribution  Applications of the Normal Distribution Probability Distributions and Statistics

2. 8.1 Distributions of Random Variables

3. Random Variable  A random variable is a rule that assigns a number to each outcome of a chance experiment.

4. Example  A coin is tossed three times.  Let the random variable X denote the number of heads that occur in the three tosses. ✦ List the outcomes of the experiment; that is, find the domain of the function X. ✦ Find the value assigned to each outcome of the experiment by the random variable X. ✦ Find the event comprising the outcomes to which a value of 2 has been assigned by X. This event is written ( X = 2 ) and is the event consisting of the outcomes in which two heads occur. Example 1, page 418

5. ExampleSolution  As discussed in Section 7.1, the set of outcomes of this experiment is given by the sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}  The table below associates the outcomes of the experiment with the corresponding values assigned to each such outcome by the random variable X:  With the aid of the table, we see that the event (X = 2) is given by the set {HHT, HTH, THH} OutcomeHHHHHTHTHTHHHTTTHTTTHTTT X32221110 Example 1, page 418

6. Applied Example: Product Reliability  A disposable flashlight is turned on until its battery runs out.  Let the random variable Z denote the length (in hours) of the life of the battery.  What values can Z assume? Solution  The values assumed by Z can be any nonnegative real numbers; that is, the possible values of Z comprise the interval 0  Z < ∞. Applied Example 3, page 419

7. Probability Distributions and Random Variables  Since the random variable associated with an experiment is related to the outcome of the experiment, we can construct a probability distribution associated with the random variable, rather than one associated with the outcomes of the experiment.  In the next several examples, we illustrate the construction of probability distributions.

8. Example  Let X denote the random variable that gives the sum of the faces that fall uppermost when two fair dice are thrown.  Find the probability distribution of X.  Solution  The values assumed by the random variable X are 2, 3, 4, …, 12, correspond to the events E 2, E 3, E 4, …, E 12.  Next, the probabilities associated with the random variable X when X assumes the values 2, 3, 4, …, 12, are precisely the probabilities P(E 2 ), P(E 3 ), P(E 4 ), …, P(E 12 ), respectively, and were computed as seen in Chapter 7.  Thus, … and so on. Example 5, page 420

9. Applied Example: Waiting Lines  The following data give the number of cars observed waiting in line at the beginning of 2-minute intervals between 3 p.m. and 5 p.m. on a given Friday at the Happy Hamburger drive-through and the corresponding frequency of occurrence.  Find the probability distribution of the random variable X, where X denotes the number of cars found waiting in line. Cars Cars012345678 Frequency Frequency29161286421 Applied Example 6, page 420

10. Applied Example: Waiting Lines Solution  Dividing each frequency number in the table by 60 (the sum of all these numbers) give the respective probabilities associated with the random variable X when X assumes the values 0, 1, 2, …, 8.  For example, Cars Cars012345678 Frequency Frequency29161286421 Applied Example 6, page 420 … and so on.

11. Applied Example: Waiting Lines Solution  The resulting probability distribution is Cars Cars012345678 Frequency Frequency29161286421 x012345678 P(X = x) P(X = x).03.15.27.20.13.10.07.03.02 Applied Example 6, page 420

12. Histograms  A probability distribution of a random variable may be exhibited graphically by means of a histogram. Examples  The histogram of the probability distribution from the last example is 012345678012345678012345678012345678.30.20.10 x

13. Histograms  A probability distribution of a random variable may be exhibited graphically by means of a histogram. Examples  The histogram of the probability distribution for the sum of the numbers of two dice is 23456789101112 6/365/364/363/362/361/36 x

14. 8.2 Expected Value

15. Average, or Mean  The average, or mean, of the n numbers x 1, x 2, …, x n is  x (read “x bar”), where

16. Applied Example: Waiting Lines  Find the average number of cars waiting in line at the Happy Burger’s drive-through at the beginning of each 2-minute interval during the period in question. Solution  Using the table above, we see that there are all together 2 + 9 + 16 + 12 + 8 + 6 + 4 + 2 + 1 = 60 numbers to be averaged.  Therefore, the required average is given by Cars Cars012345678 Frequency Frequency29161286421 Applied Example 1, page 428, refer Section 8.1

17. Expected Value of a Random Variable X  Let X denote a random variable that assumes the values x 1, x 2, …, x n with associated probabilities p 1, p 2, …, p n, respectively.  Then the expected value of X, E(X), is given by

18. Applied Example: Waiting Lines  Use the expected value formula to find the average number of cars waiting in line at the Happy Burger’s drive-through at the beginning of each 2-minute interval during the period in question. Solution  The average number of cars waiting in line is given by the expected value of X, which is given by x012345678 P(X = x) P(X = x).03.15.27.20.13.10.07.03.02 Applied Example 2, page 428, refer Section 8.1

19. Applied Example: Waiting Lines  The expected value of a random variable X is a measure of central tendency.  Geometrically, it corresponds to the point on the base of the histogram where a fulcrum will balance it perfectly: 3.1 E(X)E(X)E(X)E(X) Applied Example 2, page 428

20. Applied Example: Raffles  The Island Club is holding a fundraising raffle.  Ten thousand tickets have been sold for $2 each.  There will be a first prize of $3000, 3 second prizes of $1000 each, 5 third prizes of $500 each, and 20 consolation prizes of $100 each.  Letting X denote the net winnings (winnings less the cost of the ticket) associated with the tickets, find E(X).  Interpret your results. Applied Example 5, page 431

21. Applied Example: Raffles Solution  The values assumed by X are (0 – 2), (100 – 2), (500 – 2), (1000 – 2), and (3000 – 2).  That is –2, 98, 498, 998, and 2998, which correspond, respectively, to the value of a losing ticket, a consolation prize, a third prize, and so on.  The probability distribution of X may be calculated in the usual manner:  Using the table, we find x–2984989982998 P(X = x) P(X = x).9971.0020.0005.0003.0001 Applied Example 5, page 431

22. Applied Example: Raffles Solution  The expected value of E(X) = –.95 gives the long-run average loss (negative gain) of a holder of one ticket.  That is, if one participated regularly in such a raffle by purchasing one ticket each time, in the long-run, one may expect to lose, on average, 95 cents per raffle. Applied Example 5, page 431

23. Odds in Favor and Odds Against  If P(E) is the probability of an event E occurring, then ✦ The odds in favor of E occurring are ✦ The odds against E occurring are

24. Applied Example: Roulette  Find the odds in favor of winning a bet on red in American roulette.  What are the odds against winning a bet on red? Solution  The probability that the ball lands on red is given by  Therefore, we see that the odds in favor of winning a bet on red are Applied Example 8, page 433

25. Applied Example: Roulette  Find the odds in favor of winning a bet on red in American roulette.  What are the odds against winning a bet on red? Solution  The probability that the ball lands on red is given by  The odds against winning a bet on red are Applied Example 8, page 433

26. Probability of an Event (Given the Odds)  If the odds in favor of an event E occurring are a to b, then the probability of E occurring is

27. Example  Consider each of the following statements. ✦ “The odds that the Dodgers will win the World Series this season are 7 to 5.” ✦ “The odds that it will not rain tomorrow are 3 to 2.”  Express each of these odds as a probability of the event occurring. Solution  With a = 7 and b = 5, the probability that the Dodgers will win the World Series is Example 9, page 434

28. Example  Consider each of the following statements. ✦ “The odds that the Dodgers will win the World Series this season are 7 to 5.” ✦ “The odds that it will not rain tomorrow are 3 to 2.”  Express each of these odds as a probability of the event occurring. Solution  With a = 3 and b = 2, the probability that it will not rain tomorrow is Example 9, page 434

29. Median  The median of a group of numbers arranged in increasing or decreasing order is a.The middle number if there is an odd number of entries or b.The mean of the two middle numbers if there is an even number of entries.

30. Applied Example: Commuting Times  The times, in minutes, Susan took to go to work on nine consecutive working days were 46 42 49 40 52 48 45 43 50  What is the median of her morning commute times? Solution  Arranging the numbers in increasing order, we have 40 42 43 45 46 48 49 50 52  Here we have an odd number of entries, and the middle number that gives us the required median is 46. Applied Example 10, page 435

31. Applied Example: Commuting Times  The times, in minutes, Susan took to return from work on nine consecutive working days were 37 36 39 37 34 38 41 40  What is the median of her evening commute times? Solution  If we include the number 44 for the tenth work day and arrange the numbers in increasing order, we have 34 36 37 37 38 39 40 41  Here we have an even number of entries so we calculate the average of the two middle numbers 37 and 38 to find the required median of 37.5. Applied Example 10, page 435

32. Mode  The mode of a group of numbers is the number in the set that occurs most frequently.

33. Example  Find the mode, if there is one, of the given group of numbers. a. 1 2 3 4 6 b. 2 3 3 4 6 8 c. 2 3 3 3 4 4 4 8 Solution a.The set has no mode because there isn’t a number that occurs more frequently than the others. b.The mode is 3 because it occurs more frequently than the others. c.The modes are 3 and 4 because each number occurs three times. Example 11, page 436

34. 8.3 Variance and Standard Deviation

35. Variance of a Random Variable X  Suppose a random variable has the probability distribution and expected value E(X ) = m  Then the variance of the random variable X is x x1x1x1x1 x2x2x2x2 x3x3x3x3 · · · xnxnxnxn P(X = x) P(X = x) p1p1p1p1 p2p2p2p2 p3p3p3p3 · · · pnpnpnpn

36. Example  Find the variance of the random variable X whose probability distribution is Solution  The mean of the random variable X is given by x1234567 P(X = x) P(X = x).05.075.2.375.15.1.05 Example 1, page 442

37. Example  Find the variance of the random variable X whose probability distribution is Solution  Therefore, the variance of X is given by x1234567 P(X = x) P(X = x).05.075.2.375.15.1.05 Example 1, page 442

38. Standard Deviation of a Random Variable X  The standard deviation of a random variable X,  (pronounced “sigma”), is defined by where x 1, x 2, …, x n denote the values assumed by the random variable X and p 1 = P(X = x 1 ), p 2 = P(X = x 2 ), …, p n = P(X = x n ). p 1 = P(X = x 1 ), p 2 = P(X = x 2 ), …, p n = P(X = x n ).

39.  Let X and Y denote the random variables whose values are the weights of brand A and brand B potato chips, respectively.  Compute the means and standard deviations of X and Y and interpret your results. x15.815.916.016.116.2 P(X = x) P(X = x).1.2.4.2.1 y15.715.815.916.016.116.216.3 P(Y = y) P(Y = y).2.1.1.1.2.2.1 Applied Example: Packaging Applied Example 3, page 443

40. Applied Example: Packaging Solution  The means of X and Y are given by x15.815.916.016.116.2 P(X = x) P(X = x).1.2.4.2.1 y15.715.815.916.016.116.216.3 P(Y = y) P(Y = y).2.1.1.1.2.2.1 Applied Example 3, page 443

41. Applied Example: Packaging Solution  Therefore, the variance of X and Y are x15.815.916.016.116.2 P(X = x) P(X = x).1.2.4.2.1 y15.715.815.916.016.116.216.3 P(Y = y) P(Y = y).2.1.1.1.2.2.1 Applied Example 3, page 443

42. Applied Example: Packaging Solution  Finally, the standard deviations of X and Y are x15.815.916.016.116.2 P(X = x) P(X = x).1.2.4.2.1 y15.715.815.916.016.116.216.3 P(Y = y) P(Y = y).2.1.1.1.2.2.1 Applied Example 3, page 443

43. Applied Example: Packaging Solution  The means of X and Y are both equal to 16. ✦ Therefore, the average weight of a package of potato chips of either brand is the same.  However, the standard deviation of Y is greater than that of X. ✦ This tells us that the weights of the packages of brand B potato chips are more widely dispersed than those of brand A. x15.815.916.016.116.2 P(X = x) P(X = x).1.2.4.2.1 y15.715.815.916.016.116.216.3 P(Y = y) P(Y = y).2.1.1.1.2.2.1 Applied Example 3, page 443

44. Chebychev’s Inequality  Let X be a random variable with expected value  and standard deviation .  Then the probability that a randomly chosen outcome of the experiment lies between  – k  and  + k  is at least 1 – (1/k 2 ).  That is,

45. Applied Example: Industrial Accidents  Great Lumber Co. employs 400 workers in its mills.  It has been estimated that X, the random variable measuring the number of mill workers who have industrial accidents during a 1-year period, is distributed with a mean of 40 and a standard deviation of 6.  Use Chebychev’s Inequality to find a bound on the probability that the number of workers who will have an industrial accident over a 1-year period is between 30 and 50, inclusive. Applied Example 5, page 445

46. Applied Example: Industrial Accidents Solution  Here,  = 40 and  = 6.  We wish to estimate P(30  X  50).  To use Chebychev’s Inequality, we first determine the value of k from the equation  – k  = 30 or  + k  = 50  Since  = 40 and  = 6, we see that k satisfies 40 – 6k = 30 and 40 + 6k = 50 from which we deduce that k = 5/3. Applied Example 5, page 445

47. Applied Example: Industrial Accidents Solution  Thus, the probability that the number of mill workers who will have an industrial accident during a 1-year period is between 30 and 50 is given by that is, at least 64%. Applied Example 5, page 445

48. 8.4 The Binomial Distribution P(SFFF) = P(S)P(F)P(F)P(F) = p · q · q · q = pq 3 P(FSFF) = P(F)P(S)P(F)P(F) = q · p · q · q = pq 3 P(FFSF) = P(F)P(F)P(S)P(F) = q · q · p · q = pq 3 P(FFFS) = P(F)P(F)P(F)P(S) = q · q · q · p = pq 3

49. Binomial Experiment  A binomial experiment has the following properties: 1.The number of trials in the experiment is fixed. 2.There are two outcomes in each trial: “success” and “failure.” 3.The probability of success in each trial is the same. 4.The trials are independent of each other.

50. Example  A fair die is thrown four times. Compute the probability of obtaining exactly one 6 in the four throws. Solution  There are four trials in this experiment.  Each trial consists of throwing the die once and observing the face that lands uppermost.  We may view each trial as an experiment with two outcomes: ✦ A success ( S ) if the face that lands uppermost is a 6. ✦ A failure ( F ) if it is any of the other five numbers.  Letting p and q denote the probabilities of success and failure, respectively, of a single trial of the experiment, we find that Example 1, page 453

51. Example  A fair die is thrown four times. Compute the probability of obtaining exactly one 6 in the four throws. Solution  The trials of this experiment are independent, so we have a binomial experiment.  Using the multiplication principle, we see that the experiment has 2 4 = 16 outcomes.  The possible outcomes associated with the experiment are: 0 Successes 1 Success 2 Successes 3 Successes 4 Successes FFFFSFFFSSFFSSSFSSSS FSFFSFSFSSFS FFSFSFFSSFSS FFFSFSSFFSSS FSFS FFSS Example 1, page 453

52. 0 Success 1 Success 2 Successes 3 Successes 4 Successes FFFFSFFFSSFFSSSFSSSS FSFFSFSFSSFS FFSFSFFSSFSS FFFSFSSFFSSS FSFS FFSSExample  A fair die is thrown four times. Compute the probability of obtaining exactly one 6 in the four throws. Solution  From the table we see that the event of obtaining exactly one success in four trials is given by E = {SFFF, FSFF, FFSF, FFFS}  The probability of this event is given by P(E) = P(SFFF) + P(FSFF) + P(FFSF) + P(FFFS) 0 Successes 1 Success 2 Successes 3 Successes 4 Successes FFFFSFFFSSFFSSSFSSSS FSFFSFSFSSFS FFSFSFFSSFSS FFFSFSSFFSSS FSFS FFSS Example 1, page 453

53. Example  A fair die is thrown four times. Compute the probability of obtaining exactly one 6 in the four throws. Solution  Since the trials (throws) are independent, the probability of each possible outcome with one success is given by P(SFFF) = P(S)P(F)P(F)P(F) = p · q · q · q = pq 3 P(FSFF) = P(F)P(S)P(F)P(F) = q · p · q · q = pq 3 P(FFSF) = P(F)P(F)P(S)P(F) = q · q · p · q = pq 3 P(FFFS) = P(F)P(F)P(F)P(S) = q · q · q · p = pq 3  Therefore, the probability of obtaining exactly one 6 in four throws is Example 1, page 453

54.  In general, experiments with two outcomes are called Bernoulli trials, or binomial trials.  In a binomial experiment in which the probability of success in any trial is p, the probability of exactly x successes in n independent trials is given by Computation of Probabilities in Bernoulli Trials

55.  If we let X be the random variable that gives the number of successes in a binomial experiment, then the probability of exactly x successes in n independent trials may be written  The random variable X is called a binomial random variable, and the probability distribution of X is called a binomial distribution. Binomial Distribution

56. Example  A fair die is thrown five times.  If a 1 or a 6 lands uppermost in a trial, then the throw is considered a success.  Otherwise, the throw is considered a failure.  Find the probabilities of obtaining exactly 0, 1, 2, 3, 4, and 5 successes, in this experiment.  Using the results obtained, construct the binomial distribution for this experiment and draw the histogram associated with it. Example 2, page 455

57. ExampleSolution  This is a binomial experiment with X taking on each of the values 0, 1, 2, 3, 4, and 5 corresponding to exactly 0, 1, 2, 3, 4, and 5 successes, respectively, in five trials.  Since the die is fair, the probability of a 1 or a 6 landing uppermost in any trial is given by from which it also follows that  Finally, n = 5 since there are five trials (throws) in this experiment. Example 2, page 455

58. ExampleSolution  Using the formula for the binomial random variable, we find that the required probabilities are x012345 P(X = x) P(X = x).132 Example 2, page 455

59. ExampleSolution  Using the formula for the binomial random variable, we find that the required probabilities are x012345 P(X = x) P(X = x).132.329 Example 2, page 455

60. ExampleSolution  Using the formula for the binomial random variable, we find that the required probabilities are x012345 P(X = x) P(X = x).132.329329 Example 2, page 455

61. ExampleSolution  Using the formula for the binomial random variable, we find that the required probabilities are x012345 P(X = x) P(X = x).132.329329.165 Example 2, page 455

62. ExampleSolution  Using the formula for the binomial random variable, we find that the required probabilities are x012345 P(X = x) P(X = x).132.329329.165.041 Example 2, page 455

63. ExampleSolution  Using the formula for the binomial random variable, we find that the required probabilities are x012345 P(X = x) P(X = x).132.329329.165.041.004 Example 2, page 455

64. ExampleSolution  We can now use the probability distribution table to construct a histogram for this experiment: x012345 P(X = x) P(X = x).132.329329.165.041.004 012345012345012345012345.4.3.2.1 x Example 2, page 455

65. Applied Example: Quality Control  A division of Solaron manufactures photovoltaic cells to use in the company’s solar energy converters.  It estimates that 5% of the cells manufactured are defective.  If a random sample of 20 is selected from a large lot of cells manufactured by the company, what is the probability that it will contain at most 2 defective cells? Applied Example 5, page 457

66. Applied Example: Quality Control Solution  We may view this as a binomial experiment with n = 20 trials that correspond to 20 photovoltaic cells.  There are two possible outcomes of the experiment: defective (“success”) and non-defective (“failure”).  The probability of success in each trial is p =.05 and the probability of failure in each trial is q =.95. ✦ Since the lot from which the sample is selected is large, the removal of a few cells will not appreciably affect the percentage of defective cells in the lot in each successive trial.  The trials are independent of each other. ✦ Again, this is because of the large lot size. Applied Example 5, page 457

67. Applied Example: Quality Control Solution  Letting X denote the number of defective cells, we find that the probability of finding at most 2 defective cells in the sample of 20 is given by  Thus, approximately 92% of the sample will have at most 2 defective cells.  Equivalently, approximately 8% of the sample will contain more than 2 defective cells. Applied Example 5, page 457

68.  If X is a binomial random variable associated with a binomial experiment consisting of n trials with probability of success p and probability of failure q, then the mean (expected value), variance, and standard deviation of X are Mean, Variance, and Standard Deviation of a Random Variable

69. Applied Example: Quality Control  PAR Bearings manufactures ball bearings packaged in lots of 100 each.  The company’s quality-control department has determined that 2% of the ball bearings manufactured do not meet specifications imposed by a buyer.  Find the average number of ball bearings per package that fail to meet the buyer’s specification. Solution  Since this is a binomial experiment, the average number of ball bearings per package that fail to meet the specifications is given by the expected value of the associated binomial random variable.  Thus, we expect to find substandard ball bearings in a package of 100. Applied Example 7, page 459

70. 8.5 The Normal Distribution

71. Probability Density Functions  In this section we consider probability distributions associated with a continuous random variable: ✦ A random variable that may take on any value lying in an interval of real numbers.  Such probability distributions are called continuous probability distributions.  A continuous probability distribution is defined by a function f whose domain coincides with the interval of values taken on by the random variable associated with the experiment.  Such a function f is called the probability density function associated with the probability distribution.

72. Properties of a Probability Density Function  The properties of a probability density function are: ✦ f(x) is nonnegative for all values of x. ✦ The area of the region between the graph of f and the x-axis is equal to 1. For example: x y Area = 1

73. Properties of a Probability Density Function  Given a continuous probability distribution defined by a probability density function f, the probability that the random variable X assumes a value in an interval a < x < b is given by the area of the region between the graph of f and the x-axis, from x = a to x = b: x y Area = P(a < x < b) abababab

74. Properties of a Probability Density Function  The mean  and the standard deviation  of a continuous probability distribution have roughly the same meaning as the mean and standard deviation of a finite probability distribution.  The mean of a continuous probability distribution is a measure of the central tendency of the probability distribution, and the standard deviation measures its spread about the mean.

75. Normal Distributions  A special class of continuous probability distributions is known as normal distributions.  The normal distributions are without doubt the most important of all the probability distributions.  There are many phenomena with probability distributions that are approximately normal: ✦ For example, the heights of people, the weights of newborn infants, the IQs of college students, the actual weights of 16-ounce packages of cereal, and so on.  The normal distribution also provides us with an accurate approximation to the distributions of many random variables associated with random-sampling problems.

76. Normal Distributions  The graph of a normal distribution is bell shaped and is called a normal curve.  The curve has a peak at x = .  The curve is symmetric with respect to x = . x y 

77. Normal Distributions  The curve always lies above the x-axis but approaches the x-axis as x extends indefinitely in either direction.  The area under the curve is 1. x y Area = 1

78. Normal Distributions  For any normal curve, 68.27% of the area under the curve lies within 1 standard deviation. x y Area is.6827  –  +  

79. Normal Distributions  For any normal curve, 95.45% of the area under the curve lies within 2 standard deviations. x y Area is.9545  – 2  + 2  

80. Normal Distributions  For any normal curve, 99.73% of the area under the curve lies within 3 standard deviations. x y Area is.9973  – 3  + 3  

81. Normal Distributions  A normal distribution is completely described by the mean  and the standard deviation   : ✦ The mean  of a normal distribution determines where the center of the curve is located. x y 1111 2222 3333 4444

82. Normal Distributions  A normal distribution is completely described by the mean  and the standard deviation   : ✦ The standard deviation  of a normal distribution determines the sharpness (or flatness) of the curve. x y  1111 2222 3333 4444

83. Normal Distributions  There are infinitely many normal curves corresponding to different means  and standard deviations  .  Fortunately, any normal curve may be transformed into any other normal curve, so in the study of normal curves it is enough to single out one such particular curve for special attention.  The normal curve with mean  = 0 and standard deviation   = 1 is called the standard normal curve.  The corresponding distribution is called the standard normal distribution.  The random variable itself is called the standard normal variable and is commonly denoted by Z.

84. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(Z < 1.24). Solution  The region under the standard normal curve associated with the probability P(Z < 1.24) is z y 1.240 Example 1, page 465

85. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(Z < 1.24). Solution  Use Table 2 in Appendix B to find the area of the required region: ✦ We find the z value of 1.24 in the table by first locating the number 1.2 in the column and then locating the number 0.04 in the row, both headed by z. ✦ We then read off the number 0.8925 appearing in the body of the table, on the found row and column that correspond to z = 1.24. Example 1, page 465

86. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(Z < 1.24). Solution  Thus, the area of the required region under the curve is.8925, and we find that P(Z < 1.24) = 0.8925. z y Area is.8925 1.240 Example 1, page 465

87. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(Z > 0.5). Solution  The region under the standard normal curve associated with the probability P(Z > 0.5) is z y 0.50 Example 1, page 465

88. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(Z > 0.5). Solution  Since the standard normal curve is symmetric, the required area is equal to the area to the left of z = – 0.5, so P(Z > 0.5) = P(Z 0.5) = P(Z < – 0.5) z y Example 1, page 465

89. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(Z > 0.5). Solution  Using Table 2 in Appendix B as before to find the area of the required region we find that P(Z > 0.5) = P(Z 0.5) = P(Z < – 0.5) =.3085 z y Area is.3085 Example 1, page 465

90. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(0.24 < Z < 1.48). Solution  The region under the standard normal curve associated with the probability P(0.24 < Z < 1.48) is z y 0.24 1.48 0 P(0.24 < Z < 1.48) Example 1, page 465

91. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(0.24 < Z < 1.48). Solution  This area is obtained by subtracting the area under the curve to the left of z = 0.24 from the area under the curve to the left of z = 1.48: z y 0 P(Z < 1.48) 0.24 1.48 Example 1, page 465

92. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(0.24 < Z < 1.48). Solution  This area is obtained by subtracting the area under the curve to the left of z = 0.24 from the area under the curve to the left of z = 1.48: z y 0.24 1.48 0 P(Z < 0.24) Example 1, page 465

93. Examples  Let Z be the standard normal variable. Make a sketch of the appropriate region under the standard normal curve, and find the value of P(0.24 < Z < 1.48). Solution  Thus, the required area is given by z y 0.24 1.48 0 Area is.3358 Example 1, page 465

94. Transforming into a Standard Normal Curve  When dealing with a non-standard normal curve, it is possible to transform such a curve into a standard normal curve. If X is a normal random variable with mean  and standard deviation , then it can be transformed into the standard normal random variable Z by substituting in

95.  The area of the region under the normal curve between x = a and x = b is equal to the area of the region under the standard normal curve between  Thus, in terms of probabilities we have Transforming into a Standard Normal Curve

96.  Similarly, we have and

97. Transforming into a Standard Normal Curve  This transformation can be seen graphically as well.  The area of the region under a nonstandard normal curve between a and b is equal to the area of the region under a standard normal curve between z = (a –  )/  and z = (b –  )/   : x y abababab Nonstandard Normal Curve Same Area

98. Transforming into a Standard Normal Curve x y 0  This transformation can be seen graphically as well.  The area of the region under a nonstandard normal curve between a and b is equal to the area of the region under a standard normal curve between z = (a –  )/  and z = (b –  )/   : Standard Normal Curve Same Area

99. Example  If X is a normal random variable with  = 100 and  = 20, find the values of P(X 70), and P(75 70), and P(75 < X <110).Solution  For the case of P(X < 120), we use the formula with  = 100,  = 20, and b = 120, which gives us Example 3, page 468

100. Example  If X is a normal random variable with  = 100 and  = 20, find the values of P(X 70), and P(75 70), and P(75 < X <110).Solution  For the case of P(X > 70), we use the formula with  = 100,  = 20, and a = 70, which gives us Example 3, page 468

101. Example  If X is a normal random variable with  = 100 and  = 20, find the values of P(X 70), and P(75 70), and P(75 < X <110).Solution  Finally, for the case of P(75 < X <110), we use the formula with  = 100,  = 20, a = 75, and b = 110, which gives us Example 3, page 468

102. 8.6 Applications of the Normal Distribution

103. Applied Example: Birth Weights of Infants  The medical records of infants delivered at the Kaiser Memorial Hospital show that the infants’ birth weights in pounds are normally distributed with a mean of 7.4 and a standard deviation of 1.2.  Find the probability that an infant selected at random from among those delivered at the hospital weighed more than 9.2 pounds at birth. Applied Example 1, page 471

104. Applied Example: Birth Weights of Infants Solution  Let X be the normal random variable denoting the birth weights of infants delivered at the hospital.  Then, we can calculate the probability that an infant selected at random has a birth weight of more than 9.2 pounds by setting  = 7.4,  = 1.2, and a = 9.2 in the formula to find  Thus, the probability that an infant delivered at the hospital weighs more than 9.2 pounds is.0668. Applied Example 1, page 471

105. Approximating Binomial Distributions  One important application of the normal distribution is that it can be used to accurately approximate other continuous probability distributions.  As an example, we will see how a binomial distribution may be approximated by a suitable normal distribution.  This provides a convenient and simple solution to certain problems involving binomial distributions.

106. Approximating Binomial Distributions  Recall that a binomial distribution is a probability distribution of the form  For small values of n, the arithmetic computations may be done with relative ease. However, if n is large, then the work involved becomes overwhelming, even when tables of P(X = x) are available.

107. Approximating Binomial Distributions  To see how a normal distribution can help in such situations, consider a coin-tossing experiment.  Suppose a fair coin is tossed 20 times and we wish to compute the probability of obtaining 10 or more heads.  The solution to this problem may be obtained, of course, by laboriously computing  As an alternative solution, let’s begin by interpreting the solution in terms of finding the areas of rectangles in the histogram.

108. Approximating Binomial Distributions  We may calculate the probability of obtaining exactly x heads in 20 coin tosses with the formula  The results lead to the binomial distribution shown in the table to the right. x P(X = x) 0.0000 1.0000 2.0002 3.0011 4.0046 5.0148 6.0370 7.0739 8.1201 9.1602 10.1762 11.1602 12.1201  20.0000

109. Approximating Binomial Distributions  Using the data from the table, we may construct a histogram for the distribution:.20.15.10.05 0 12345678910 12 14 16 18 20 12345678910 12 14 16 18 20 x

110. Approximating Binomial Distributions  The probability of obtaining 10 or more heads in 20 coin tosses is equal to the sum of the areas of the blue shaded rectangles of the histogram of the binomial distribution:.20.15.10.05 0 12345678910 12 14 16 18 20 12345678910 12 14 16 18 20 x

111. Approximating Binomial Distributions  Note that the shape of the histogram suggests that the binomial distribution under consideration may be approximated by a suitable normal distribution..20.15.10.05 0 12345678910 12 14 16 18 20 12345678910 12 14 16 18 20 x

112. Approximating Binomial Distributions  The mean and standard deviation of the binomial distribution in this problem are given, respectively, by  Thus, we should choose a normal curve for this purpose with a mean of 10 and a standard deviation of 2.24.

113. Approximating Binomial Distributions  Superimposing on the histogram a normal curve with a mean of 10 and a standard deviation of 2.24 clearly gives us a good fit:.20.15.10.05 0 12345678910 12 14 16 18 20 12345678910 12 14 16 18 20 x

114. Approximating Binomial Distributions  The good fit suggests that the sum of the areas of the rectangles representing P(X = x), the probability of obtaining 10 or more heads in 20 coin tosses, may be approximated by the area of an appropriate region under the normal curve..20.15.10.05 0 12345678910 12 14 16 18 20 12345678910 12 14 16 18 20 x

115. Approximating Binomial Distributions  To determine this region, note that the base of the portion of the histogram representing the required probability extends from x = 9.5 on, since the base of the leftmost rectangle is centered on 10:.20.15.10.05 0 12345678910 12 14 16 18 20 12345678910 12 14 16 18 20 x x  10 9.5

116. Approximating Binomial Distributions  Therefore, the required region under the normal curve should also have x  9.5.  Letting Y denote the continuous normal variable, we obtain

117. Approximating Binomial Distributions  The exact value can be found by computing in the usual (time-consuming) fashion.  Using this method yields a probability of.5881, which is not very different from the approximation of.5871 obtained using the normal distribution.

118.  Suppose we are given a binomial distribution associated with a binomial experiment involving n trials, each with a probability of success p and a probability of failure q.  Then, if n is large and p is not close to 0 or 1, the binomial distribution may be approximated by a normal distribution with Theorem 1

119. Applied Example: Quality Control  An automobile manufacturer receives the microprocessors used to regulate fuel consumption in its automobiles in shipments of 1000 each from a certain supplier.  It has been estimated that, on the average, 1% of the microprocessors manufactured by the supplier are defective.  Determine the probability that more than 20 of the microprocessors in a single shipment are defective. Applied Example 4, page 475

120. Applied Example: Quality Control Solution  Let X denote the number of defective microprocessors in a single shipment.  Then X has a binomial distribution with n = 1000, p =.01, and q =.99, so Applied Example 4, page 475

121. Applied Example: Quality Control Solution  Approximating the binomial distribution by a normal distribution with a mean of 10 and a standard deviation of 3.15, we find that the probability that more than 20 microprocessors in a shipment are defective is given by  Thus, approximately 0.04% of the shipments with 1000 microprocessors each will have more than 20 defective units. Applied Example 4, page 475

122. End of Chapter