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# Chapter 5: Normal Probability Distributions - Solutions

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 .

Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate. Your answers may vary slightly. 5.2 Normal Distributions: Finding Probabilities

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### Transcription of 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 .

2 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 . 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.

3 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? 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?

4 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. 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.

5 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. 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.

6 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 . 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 ).

7 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. 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).

8 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. 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=.

9 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.

10 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.