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 xz 15,000 miles : 1.94 1290 Pz( 1.94) 1 0.9738 0.0262
X 8,000 12,494 For xz 8,000 miles : 3.48 1290 Pz( 3.48) 0.0003
X 6,000 12,494 For xz 6,000 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. Thus, z= -1.645. xz x 1.645 18 122.6 92.99 or 93 Therefore, a cutoff score of 93 will make a student eligible for the program.
23. Used Car Prices an automobile dealer finds that the average price of a previously owned vehicle is $8256. He decides to sell cars that will appeal to the middle 60% of the market in terms of price. Find the maximum and minimum prices of the cars the dealer will sell. The standard deviation is $1150, and the variable is normally distributed.
$8256, $1150 . The middle 60% means that 30% of the area will be on either side of the mean. The corresponding z score is found using area = 0.20. The z scores are 0.84.
Min z * 0.84*1150 8256 $7290
Max z* 0.84*1150 8256 $9222
26. High School Competency Test a mandatory competency test for high school sophomores has a normal distribution with a mean of 400 and a standard deviation of 100. a) The top 3% of students receive $500. What is the minimum score you would need to receive this award?
400, 100 1 0.03 0.97 z 1.88 Min z * 1.88*100 400 588
588 is minimum score to receive the award. b) The bottom 1.5% of students must go to summer school. What is the minimum score you would need to stay out of this group?
400, 100
For the bottom 1.5%, the area is 0.015. z 2.17
Min z * 2.17*100 400 183
183 is the minimum score to avoid summer school.
28. Bottled Drinking Water Americans drank an average of 23.2 gallons of bottled water per capita in 2004. If the standard deviation is 2.7 gallons and the variable is normally distributed, find the probability that a randomly selected American drank more than 25 gallons of bottled water. What is the probability that the selected person drank between 18 and 26 gallons?
23.2, 2.7
For x 25 gallons: 25 23.2 z 0.67; the corresponding area=0.7486 2.7 0.67 Pz( 0.67) 1 0.7486 0.2514
For 18x 26gallons: 18 23.2 z 1.93; the corresponding area=0.0268 2.7
26 23.2 z 1.04; the corresponding area=0.8508 2.7 Px(18 1.04) 0.8508 0.0268 = 0.824 -1.93 1.04
30. Security Officer Stress Tolerance to qualify for security officers’ training, recruits is tested for stress tolerance. The scores are normally distributed, with a mean of 62 and a standard deviation of 8. If only the top 15% of recruits are selected, find the cutoff score.
62, 8 1 0.15 0.85
The corresponding z score is 1.04 Cutoff z 1.04 8 62 70.32
6.3 #9, 13, 15, 17, 21, 23
9. College Costs The mean undergraduate cost for tuition, fees, room, and board for four-year institutions was $26,489 for the 2004–2005 academic year. Suppose that σ =$3204 and that 36 four-year institutions are randomly selected. Find the probability that the sample mean cost for these 36 schools is a. Less than $25,000
µ=26,489, n=36, σ= 3204, P( x < 25.000) x x 25,000 26,489 Z 2.79 3204 n 36 -2.97 P( z < -2.79)=0.0026 b. Greater than $26,000
µ=26,489, n=36, σ= 3204, P( x >26.000)
26,000 26,489 0.92 3204 36 -0.92 P( z > -0.92)=1-0.1788 = 0.8212 c. Between $24,000 and $26,000.
µ=26,489, n=36, σ= 3204, P($24,000 < <$26.000)
24,000 26,489 4.66 3204 36 -4.66 -0.92 26,000 26,489 0.92 3204 36
P(-4.66< <-0.92)= 0.1788 -0.0001 = 0.1787
13. Fuel Efficiency for U.S. Light Vehicles The average fuel efficiency of U.S. light vehicles (cars, SUVs, minivans, vans, and light trucks) for 2005 was 21 mpg. If the standard deviation of the population was 2.9 and the gas ratings were normally distributed, what is the probability that the mean mpg for a random sample of 25 light vehicles is under 20? Between 20 and 25?
For x< 20 mpg: x x 20 21 Z x 1.72 2.9 x n 25 -1.72 P(z<- 1.72)= 0.0427 or 4.27%
For 20< x < 25;
25 21 6.90 2.9 25
-172 6.90
Z=6.90; the corresponding area=0.9999 Z=-1.72; the corresponding area=0.0427 P(-1.72< x <6.90)= 0.9999-0.0427=0.9572 or 95.72%
15. Sodium in Frozen Food The average number of milligrams (mg) of sodium in a certain brand of low- salt microwave frozen dinners is 660mg, and the standard deviation is 35 mg. Assume the variable is normally distributed. a. If a single is selected, find the probability that the sodium content will be more than 670 mg.
µ=660, σ= 35, P( x < 670)
670 660 10 Z 0.285 0.29 35 35
Pz( 0.29) 1 0.6141 0.3859or 38.59%
b. If a sample of 10 dinners is selected, find the probability that the mean of the sample will be larger than 670 mg.
µ=660, n=10, σ= 35, P( x <670)
670 600 10 10 Z 0.902 0.90 35 35 11.08 10 3.16 Pz( 0.90) 1 0.8159 0.1841 18.41% c) Why is the probability for part a greater than that for part b?
Because the individual values are more variable than means.
17. Water Use The Old Farmer’s Almanac reports that the average person uses 123 gallons of water daily. If the standard deviation is 21 gallons, find the probability that the mean of a randomly selected sample of 15 people will be between 120 and 126 gallons. Assume the variable is normally distributed.
µ=123, n=15, σ= 21, normally distributed
P(120< x <126)
120 123 3 3 Z 0.55 21 21 5.43 15 3.87
126 123 3 3 Z 0.55 21 21 5.43 120 123 126 15 3.87
Pz( 0.55 0.55) 0.7088 0.2912
0.4176 41.76%
21. Time to Complete an Exam The average time it takes a group of adults to complete a certain achievement test is 46.2 minutes. The standard deviation is 8 minutes. Assume the variable is normally distributed. a. Find the probability that a randomly selected adult will complete the test in less than 43 minutes.
µ=46.2, σ= 8, P( <43) normally distributed
43 46.2 3.2 Z 0.4 8 8
Pz( 0.4) 0.3446 or 34.46%
b. Find the probability that, if 50 randomly selected adults take the test, the mean time it takes the group to complete the test will be less than 43 minutes.
µ=46.2, σ= 8, n=50 P( x <43) 43 46.2 3.2 3.2 z 2.83 8 8 1.13 50 7.07
23. Cholesterol Content The average cholesterol content of a certain brand of eggs is 215 milligrams, and the standard deviation is 15 milligrams. Assume the variable is normally distributed. a. If a single egg is selected, find the probability that the cholesterol content will be greater than 220 milligrams.
µ=215, σ= 15, P( x >220), normally distributed
220 215 5 Z 0.33 15 15
Pz( 0.33) 1 0.6293 0.3707 or 37.07%
b. If a sample of 25 eggs is selected, find the probability that the mean of the sample will be larger than 220 milligrams.
µ=215, n=25, σ= 15, P( x >220)
220 215 5 5 z 1.67 15 15 3 25 5
Pz( 1.67) 1 0.9525 0.0475 or 4.75%
Ch 6.4 p.344 #3a, b, c, d, 7, 8, 11
3. Check each binomial distribution to see whether it can be approximated by a normal distribution (i.e., are np ≥ 5 and nq ≥ 5?). a. n=20, p=0.5 q = 1–q = 1 – 0.5 = 0.5 np = (20)(0.5) =10 nq = (20)(0.5) = 10
Yes, because np ≥ 5 and nq ≥ 5. b. n=10, p=0.6 q = 1–q = 1 – 0.6 = 0.4 np = (10)(0.6) =6 nq = (10)(0.4) = 4
No because np ≥ 5 and nq ≤ 5.
c. n=40, p=0.9 q = 1–q = 1 – 0.9 = 0.1 np = (40)(0.9) =36 nq = (40)(0.1) = 4
No because np ≥ 5 and nq ≥ 5.
d. n=50, p=0.2 q = 1–q = 1 – 0.2 = 0.8 np = (50)(0.2) =10 nq = (50)(0.8) = 40
Yes, because np ≥ 5 and nq ≥ 5..
7. Percentage of Americans Who Have Some College Education The percentage of Americans 25 years or older who have at least some college education is 53.1%. In a random sample of 300 Americans 25 years old or older, what is the probability that more than 175 have at least some college education?
n=300, p=0.531 q = 1–q = 1 – 0.531 = 0.469
Since np = (300)(0.531) = 159.3 and nq = (300)(0.469) = 140.7, the normal distribution can be used to approximate the binomial distribution (np ≥ 5 and nq ≥ 5).
µ = np = (300)( 0.531) = 159.3
σ = npq = (300)(0.531)(0.469) 8.64
For binomial distribution, P(X >175)
For normal distribution, P(X >175 + 0.5) = P(X > 175.5) 153.9 175.5 175.5 159.3 The z value is z = 1.88 8.64
From table E (Normal Distribution ), z=1.88 → 0.9699
1 – 0.9699 = 0.0301 or 3.01%
8. Household Computers According to recent surveys, 60% of households have personal computers. If a random sample of 180 households is selected, what is the probability that more than 60 but fewer than 100 have a personal computer? n=180, p=0.60 q = 1–q = 1 – 0.60 = 0.40
Since np = (180)(0.60) = 108 and nq = (180)(0.40) = 72, the normal distribution can be used to approximate the binomial distribution (np ≥ 5 and nq ≥ 5).
µ = np = (180)(0.60) = 108
σ = npq = (180)(0.60)(0.40) 6.57
For binomial distribution, Px(60 100)
For normal distribution, P(60.5 < X < 99.5)
= P(60.5 < X < 99.5)
99.5 108 z = 1.29 area=0.0985 6.57
60.5 108 z = 7.23 area=0.0001 6.57
P(60.5 < X < 99.5)= P(-7.23 < z< -1.29)=0.0985-0.0001=0.0984 or 9.84%
0.4955 + 0.4945 = 0.99 or 0.99%
11. Elementary School Teachers Women comprise 80.3% of all elementary school teachers. In a random sample of 300 elementary teachers, what is the probability that more than three-fourths are women? n=300, p=0.803 q = 1–q = 1 – 0.803= 0.197
Since np = (300)(0.803) = 240.9 and nq = (300)(0.197) = 59.1, the normal distribution can be used to approximate the binomial distribution (np ≥ 5 and nq ≥ 5).
µ = np= (300)(0.803) = 240.9
σ = npq = (300)(0.803)(0.197) 6.89
For binomial distribution, P(X >225) ¾ (300)=225
For normal distribution, P(X >255 + 0.5) = P(X > 255.5)
240.9 255.5 225.5 240.9 The z value is z = 2.24 area=0.0125 6.89
P(z > 225.5)=P(X > -2.24)=1- 0.0125 =0.9875 or 98.75%