Chapter 5: Normal Probability Distributions - Solutions

1 Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. Your answers may vary slightly. Normal Distributions : Finding Probabilities If you are given that a random variable X has a Normal distribution, finding probabilities corresponds to finding the area between the standard Normal curve and the x-axis, using the table of z-scores. The mean (expected value) and standard deviation should be given in the problem. For the Probability that X < b, convert b into a z-score using b . z=.. and use the table to find the area to the left of the z-value. For the Probability that X > a, convert a into a z-score using a . z=.. and use the table to find the area to the right of the z-score. For the Probability that a < X < b (X is between two numbers, a and b), convert a and b into z-scores using a b.

2 Z= and z =.. and use the table to find the area between the two z-values. 1. The average speed of vehicles traveling on a stretch of highway is 67 miles per hour with a standard deviation of miles per hour. A vehicle is selected at random. a. What is the Probability that it is violating the 70 mile per hour speed limit? Assume that the speeds are normally distributed. Solution: The random variable X is speed . We are told that X has a Normal distribution. The mean = 67 . The standard deviation = . We are looking for the Probability of the event that X > 70 . 1. Step 1: convert 70 into a z-score: 70 67. z= Step 2: Find the appropriate area between the Normal curve and the axis using the table: The table contains cumulative areas (to the left of the z-value). The area cor- responding to a z-score of in the table is Since we are interested in X > 70, we need the area to the right of the z-score, thus P (X > 70) 1 (a) What is the Probability that a randomly selected vehicle is not violating the speed limit?

3 The z-score is the same: We are interested in P (X 70), thus the area to the left of this z-score can be read directly off the table: OR, using complements and the answer to part a, P (X 70) = 1 P (X > 70) 1 (b) What is the Probability that a randomly selected vehicle is traveling under 50. miles per hour? We are interested in P (X < 50). The z-score is 50 67 17. z= = The area needed is to the left of this z-score. Notice that is not even on the table, and the lowest z-score is , with a corresponding area of For z-scores less than , the area is even less than and very close to 0, and we may assume is approximately 0. (The more accurate answer is about : about 1 in million Probability .). (c) What is the Probability that a randomly selected vehicle is traveling between 50.

4 And 70 miles per hour? The z-score for 70 is with a corresponding area of The z-score for 50. is with a corresponding area of about 0. Thus, we subtract: P (50 < X < 70) 0 2. Practice Problem: A customer calling a call center spends an average of 45 minutes on hold during the peak season, with a standard deviation of 12 minutes. Suppose these times are normally distributed. Find the Probability that the customer will be on hold for each interval of times: a. More than 54 minutes. Let X be the number of minutes the customer spends on hold. We want P (X > 54). The mean is = 45 and standard deviation is = 12. The z-score is 54 45 9. z= = = 12 12. The corresponding area is For P (X > 54), the area to the right is needed. Thus, P (X > 54) 1 = b. Less than 24 minutes.

5 We want P (X < 24). The z-score is 24 45 21. z= = = 12 12. The corresponding area is For P (X < 24), the area to the left is needed. Thus, P (X < 24) c. Between 24 and 54 minutes. The z-score for 24 is with a corresponding area of , and the z-score for 54 is with a corresponding area of Thus, P (24 < X < 54) = d. More than 39 minutes. We need P (X > 39). The z-score for 39 is 39 45. z= = 12. The corresponding area is We need the area to the right of the z-score. Thus, P (X > 39) 1 = 3. Normal Distributions : Finding Values Now the process from will be reversed. Starting with a Probability , you will find a corresponding z-score. The same table will be used, but you will search the center of the table to find the Probability first, and then determine the z-score that corresponds to that Probability .

6 To make this easier, first draw a picture. 2. Find the indicated z-scores. Draw a picture and include a short explanation a. The z-score that corresponds to a cumulative area of (the cumulative area is the area to the left of the z-score). Look for the given area in the table and find the corresponding z-score: b. The z-score that corresponds to of the distribution's area to its right. The table lists the cumulative area: to the left of the z-score. Thus, the z-score needed corresponds to a left area of 1 = This z-score is c. The z-score that corresponds to of the distribution's area to its right. First convert into a Probability (area): The z-score needed corre- sponds to a left area of 1 = This z-score is d. The z-score that corresponds to the 90th percentile (P90 ) of the distribution's area.

7 convert 90% into a Probability (area) first: Even though this exact area is not in the table, pick the closest areas. The desired z-score is between and (You may use either of these or average them). Practice Problem: a. The z-score that corresponds to a cumulative area of From the table: z = b. The z-score that corresponds to of the distribution's area to its right. Find the z-score corresponding to area 1 = This is c. The z-score that corresponds to of the distribution's area to its right. convert into a Probability : , and find the z-score corresponding to area 1 = This is d. The z-score that corresponds to the 30th percentile (P30 ) of the distribution's area. convert 30% into a Probability : , and find the z-score corresponding to this area: between and 4.

8 Transforming a z-score into a data value Given a z-score, it can be converted back into a data value by solving for x in the equation x . z=.. Given z, to find x, use the formula x = + z . Procedure: Area z-score data value. 3. Scores for the California Police Officer Standards and Training test are normally dis- tributed, with a mean of 50 and a standard deviation of 10. a. An agency will only hire applicants with scores in the top 10%. What is the lowest score you can earn and still be eligible to be hired by the agency? The mean is = 50 and standard deviation is = 10. The top 10% corresponds to the 90th percentile. The corresponding z-score was found earlier, which is about Using the formula, this corresponds to a test score of x = + z 50 + (10) 63. b. Those officers scoring below the 20th percentile are sent to undergo additional train- ing.

9 What is the minimum score needed to avoid this training? The 20th percentile corresponds to a cumulative area of The closest z-scores are and We can use the average z-score This corresponds to a test score of x = + z 50 + ( )(10) 42. Practice Problem: The length of time employees have worked at a particular company is normally distributed with mean years and standard deviation years. a. If the lowest 10% of employees in seniority are to be layed-off in a cutback, what is the maximum length of time that an employee could have worked and still be laid off? The 10th percentile corresponds to a cumulative area of The closest z-scores are and We can use the average z-score This corresponds to the length of time worked x = + z + ( )( ) years 5. b. If the highest 10% of employees in seniority are to be promoted, what is the minimum length of time that an employee could have worked and still be promoted?

10 The 90th percentile corresponds to a cumulative area of The closest z-scores are and We can use the average z-score This corresponds to the length of time worked x = + z + ( )( ) years 6. Sampling Distributions and The Central Limit Theorem Given: i) a (large) population, ii) a numerical characteristic associated with each member of the population, iii) the population mean and population standard deviation for this characteristic, You: i) take a simple random sample of 100 members of the population and calculate the mean and standard deviation. ii) repeat taking simple random samples of 100 members several times and calculate the mean and standard deviation each time. The sample distribution is denoted by x . In general, you cannot expect that the mean you obtain for each sample of 100 to be equal to , but Theorem (Central Limit Theorem).