Ch. 6.1 #7-49 odd The area is found by looking up z= 0.75 in Table E and subtracting 0.5. Area = 0.7734- 0.5= 0.2734 The area is found by looking up z= 2.07 in Table E and subtracting from 0.5. Area = 0.5- 0.0192 = 0.4808 The area is found by looking up z=0.23 in Table E and subtracting it from 1 Area =1-0.5910= 0.4090 The area is found by looking up z= 1.43 in Table E. Area = 0.0764 15) Between z = 1.05 and z= 1.78 The area is found by looking up the values 1.05 and 1.78 in Table E and subtracting the areas. Area= 0.9625– 0.8531 = 0.1094. 0.1094 Z=1.78 Z=1.05 58 The area is found by looking up the values z=1.56 and z=1.83 in Table E and subtracting the areas. Area = 0.0594 - 0.0336 = 0.0258 19) Between z = -1.53 and z= - 2.08 The area is found by looking up the values z=1.53 and z= 2.08 in Table E and subtracting the areas. Area = 0.0630- 0.0188 = 0.0442 . The area is found by looking up z= 2.11 in Table E. Area = 0.9826 23) To the right of z=-0.25 The area is found by looking up z= 0.25 in Table E and subtracting it from 1. Area= 1- 0.4013 = 0.5987 For z = -0.44, the area is 0.3300. For z = 1.92, the area is 1 - 0.9726 = 0.0274 Area = 0.3300 + 0.0274 = 0.3574 Area = 0.7486 - 0.5 = 0.2486 Area = 0.5 - 0.0582 = 0.4418 Area =1 - 0.9977 = 0.0023 Area = 0.1131 Area= 0.9591 - 0.0069 = 0.9522 Area =0.9985 - 0.9279 = 0.0706 Area = 0.9222 For exercises 40 through 45, find the z value that corresponds to the given area. 41) Since the z score is on the left side of 0, use the negative z table. Areas in the negative z table are in the tail, so we will use 0.5 - 0.4175 = 0.0825 as the area. The closest z score corresponding to an area of 0.0825 is z=- 1.39. 43) Z=- 2.08, found by using the negative z table. 45) Use the negative z table and 1 - 0.8962 = 0.1038 for the area. The z score is z= -1.26. 47) Find the z values to the left of the mean so that a) 98.87% of the area under the distribution curve lies to the right of it. Using the negative z table, area = 1 - 0.9887 = 0.0113. Hence Z=- 2.28. b) 82.12% of the area under the distribution curve lies to the right of it. Using the negative z table, area = 1 – 0.8212 = 0.1788. Hence Z=- 0.92. c) 60.64% of the area under the distribution curve lies to the right of it. Using the negative z table, area = 1 – 0.6064 = 0.3936. Hence Z=- 0.27. 49) Find two z values, one positive and one negative, so that the areas in the two tails total the following values. a) 5% For total area = 0.05, there will be area = 0.025 in each tail. The z scores are 1.96. b) 10% For total area =0.10, there will be area = 0.05 in each tail. The z scores are 1.645. c) 1% For total area = 0.01, there will be area = 0.005 in each tail. The z scores are z= 2.58. Section 6-2 # 1, 2, 9, 11, 15, 21, 22, 23, 26, 28, 30 1. Admission Charges for Movies The average admission charge for a movie is $5.81. If the distribution of movie admission charges is approximately normal with a standard deviation of $0.81, what is the probability that a randomly selected admission charge is less than $3.50? X $3.50 $5.81 z 2.85 $0.81 then look up z = -2.85 in Table E and you get that the area = 0.0022 or 0.22% Pz( 2.85) 0.0022 or 0.22% -2.85 0 2. Teachers’ Salaries The average annual salary for all U.S. teachers is $47,750. Assume that the distribution is normal and the standard deviation is $5680. Find the probability that a randomly selected teacher earns a. Between $35,000 and $45,000 a year X 35,000 47,750 z 2.24 5680 45,000 47,750 z 0.48 5680 Pz( 2.24 0.62) 0.3156 0.0125 0.3031 or 30.31% b. More than $40,000 a year X 40,000 47,750 z 1.36 5680 Pz( 1.36) 1 0.0869 0.9131 or 91.31% c. If you were applying for a teaching position and were offered $31,000 a year, how would you feel (based on this information)? Not too happy! It's really at the bottom of the heap! X 31,000 47,750 z 2.95 5680 Pz( 2.95) 0.0016 Only 0.16% of salaries are below $31,000. 9. Miles Driven Annually The mean number of miles driven per vehicle annually in the United States is 12,494 miles. Choose a randomly selected vehicle, and assume the annual mileage is normally distributed with a standard deviation of 1290 miles. What is the probability that the vehicle was driven more than 15,000 miles? Less than 8000 miles? Would you buy a vehicle if you had been told that it had been driven less than 6000 miles in the past year? X 15,000 12,494 For xz15,000 miles : 1.94 1290 Pz( 1.94) 1 0.9738 0.0262 X 8,000 12,494 For xz8,000 miles : 3.48 1290 Pz( 3.48) 0.0003 X 6,000 12,494 For xz6,000 miles : 5.03 1290 Pz( 5.03) 0.0001 Maybe it would be good to know why it had only been driven less than 6000 miles. 11. Credit Card Debt The average credit card debt for college seniors is $3262. If the debt is normally distributed with a standard deviation of $1100, find these probabilities. a) That the senior owes at least $1000 X 1000 3262 2262 z 2.06 1100 1100 Pz( 2.06) 1 0.0197 0.9803 or98.03% -2.06 0 b) That the senior owes more than $4000 X 4000 3262 738 z 0.67 1100 1100 Pz( 0.67) 1 0.7486 0.2514 or 25.14% 0 0.67 c) That the senior owes between $3000 and $4000 X 3000 3262 262 z 0.24 1100 1100 X 4000 3262 738 z 0.67 1100 1100 Pz( 0.24 0.67) 0.7486 0.4052 0.3434 or 34.34% -0.24 0.67 15. Waiting to Be Seated The average waiting time to be seated for dinner at a popular restaurant is 23.5 minutes, with a standard deviation of 3.6 minutes. Assume the variable is normally distributed. When a patron arrives at the restaurant for dinner, find the probability that the patron will have to wait the following time. a) Between 15 and 22 minutes X 15 23.5 8.5 z 2.36 3.6 3.6 X 22 23.5 1.5 z 0.42 3.6 3.6 Pz( 2.36 0.42) 0.3372 0.0091 0.3281 -2.36 -0.42 b) Less than 18 minutes or more than 25 minutes X 18 23.5 5.5 z 1.53 3.6 3.6 X 25 23.5 1.5 z 0.42 3.6 3.6 P( z 1.53 or z 0.42) 0.0630 (1 0.6628) 0.063 0.3372 0.4002 -1.53 0.42 c) Is it likely that a person will be seated in less than 15 minutes? Pz( 2.36) 0.0091 Since the probability is small, it is not likely that a person would be seated in less than 15 minutes. 21. Cost of Personal Computers The average price of a personal computer (PC) is $949. If the computer prices are approximately normally distributed and $100 , what is the probability that a randomly selected PC costs more than $1200? The least expensive 10% of personal computers cost less than what amount? Let X be the amount a least expensive personal computer can cost. X 1200 949 251 z 2.51 100 100 Pz( 2.51) 1 0.9940 0.006 or 0.06% 2.51 For the least expensive 10%, the corresponding z-value is z = - 1.28. x 949 1.28 100 Area is .10 x 1.28(100) 949 $821 Z= -1.28 22. Reading Improvement Program To help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom 5% of the students in the district, based on the scores on a reading achievement exam. If the average score for the students in the district is 122.6, find the cutoff score that will make a student eligible for the program. The standard deviation is 18. Assume the variable is normally distributed. 122.6, 18 The bottom 5% (area) is in the left tail of the normal curve. The corresponding z score is found using area = 0.05.