Download Ch. 6.3 #7-49 odd To find the area between the z=0 and z=0.75

Ch. 6.3 #7-49 odd
To find the area between the z=0 and z=0.75, look up the area in table E corresponding to
z=0.75. Therefore, area that corresponds to this z value is 0.2734.
To find the area between the z=0 and z=-2.07, look up the area in the table E corresponding to
z=2.07 because the normal distribution is symmetric about the mean, the area to the left of the
mean is the same as the area to the right of the mean. Therefore, the area that corresponds to this
value is 0.4808.
Since this area is a tail area, find the area between z=0 and z=0.23 and subtract it from 0.5000.
Hence, 0.5000 - 0.0910 = 0.4090.
Table E gives us the area for positive z values. Because of the symmetric property of normal
distribution, the area to the left of -1.43 is the same as the area to the right of +1.43. Now, find
the area between z=0 and z=1.43 which is 0.4236 and subtract it from 0.5000. Therefore, the area
is equal to 0.0764.
To find the area between z=0.79 and z=1.28, first look up the area from z=0 to z=1.28 which is
0.3997, and the area from z=0 to z=0.79 which is 0.2852, then subtract the smaller area from the
larger area. Hence, 0.3997 – 0.2852 = 0.1145.
First find the area between z=0 and z=1.83 which is 0.4664, then find the area between z=0 and
z=1.56 which is 0.4406, and subtract the smaller area from the larger area to get the answer.
Hence, 0.4664 – 0.4406 = 0.0258.
Since the two area are on opposite sides of z=0, one must find both areas and add them. First find
the area between z=0 and z=2.47 which is 0.4932, then find the area between z=0 and z=-1.03
which is 0.3485, and add them. Hence, 0.4932 + 0.3485 = 0.8417.
Because Table E only gives us the area between z=0 and z=2.11, we must add 0.5000 to the table
area, because 0.5000 is the total area that rests to the left of z=0. Hence, 0.4826 + 0.5000 =
0.9826.
Find the area between z=0 and z=-0.15 which is 0.0596, then add 0.5000 to it. Hence, 0.0596 +
0.5000 = 0.5596.
The area to the right of z=1.92 is 0.5000 - 0.4726 = 0.0274, the area to the left of z=-0.44 is
0.5000 – 0.1700 = 0.3300. The total area, then, is 0.0274 + 0.3300 = 0.3574.
Look up the area in table E corresponding to z=0.67. It is 0.2486.
Look up in table E for the area corresponding to the z=1.57. We do this since the area to the
right of the mean is the same as the area to the left of the mean because the symmetry of the
normal distribution. Thus, the area that corresponds to this z value is 0.4418.
Use table E to look up for the area that corresponds to z=2.83. Once you have found this area
which is 0.4977, since we are interested on the right hand side of z=2.83 subtract 0.4977 from
0.5000. Thus, the area that corresponds to z>2.83 is 0.5000-0.4977=0.0023
To find the area that corresponds to z=-1.21, look up the area that corresponds to z=1.21 because
we know that a normal distributed graph is symmetric about the mean (the area to the right of the
mean is the same as the area to the left of the mean). The area that corresponds to z=1.21 is
0.3869. We are interested in the area to the left of z=-1.21. For this reason, subtract 0.3869 from
0.5000.
Then P(z<-1.21) = 0.5000-0.3869 = 0.1131
Use table E. First find the area that corresponds to z=1.74 which is 0.4591, then find the area that
corresponds to z=2.46 (by symmetry) which is 0.4931. Since one of the z values lies above the
mean and the other below the mean, we add their corresponding areas 0.4591+0.4931=0.9522.
Then, the P (2.46  z  1.74)  0.9522
Use table E. Find the area corresponding to the z=1.46 which is 0.4279 and the area
corresponding to z=2.97 which is 0.4985. Since both z values are on the same side of the mean,
we subtract the smaller area from the larger. Then
P (1.46  z  2.97)  0.4985  0.4279  0.0706 .
Using table E, find the area that corresponds to z=1.42 which is 0.4222, and since we are
interested in the area to the left hand side of z=1.42 we add its area to 0.5000. Then
P ( z  1.42)  0.5000  0.4222  0.9222
For exercises 40 through 45, find the z value that corresponds to the given area.
41)
Look up for the z value that corresponds to area of 0.4738. The z value corresponding to this
area is -1.94
43)
Since the given area is a left tail area, it means that to get the corresponding z value first we need
to subtract 0.0166 from 0.5000. Hence 0.5000-0.0166=0.4834. Now use table E to look up for
the z value whose area is 0.4834. For this area z=2.13, however since this is a left tail area
z=-2.13.
45) For this exercise, the area of 0.8962 lies to the right of the z value. To find the z value, first
subtract 0.5000 from 0.8962. 0.8962-0.5000=0.3962. Now look up table E and find the z value
that corresponds to this area. For this area z= 1.26, but since the z value is below the mean then
it has to be negative, then 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.
Similarly to exercise # 45, first subtract 0.5000 from 0.9887.
0.9887-0.5000=0.4887. Now, find the z value whose area is 0.4887 which is z=2.28, but since
the z values lies below the mean then it has to be negative, thus z= -2.28
b) 82.12% of the area under the distribution curve lies to the right of it.
0.8212-0.5000=0.3212
z-value corresponding to area of 0.3212 is 0.92, but since the z value lies below the mean, then it
becomes z=-0.92.
c) 60.64% of the area under the distribution curve lies to the right of it.
0.6064-0.5000=0.1064
The z value whose area is 0.1064 is 0.27, thus the z value to the left of the mean is 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%
0.05/2=0.025
Therefore each tail has area of 0.025. Use table E to find the z value whose area to the right of it
is 0.025. So subtract 0.025 from 0.5000, 0.5000-0.025=0.475. The z value whose area is 0.475
is 1.96. Now by symmetry, the z value above the mean is z=1.96, and the z value below the
mean is z=-1.96.
b) 10%
.010/2=0.05
Each tail has an area of 0.05. Similarly as exercise (a), subtract 0.05 from 0.5000.
0.5000-0.05=0.4500
The z value whose area is 0.4500 is 1.65. Thus by symmetry, the z value above the mean is
z=1.65, and the z value below the mean is z=-1.65.
c) 1%
0.01/2=0.005
Each tail has an area of 0.005.
0.5000-0.005=0.495
The z value whose area is 0.495 is 2.58. Now by symmetry, the z value above the mean is
z=2.58, and the z value below the mean is z=-2.58.
Section 6-4 Pg. 312 Exercises 1, 2, 9, 11, 15, 21, 22, 23, 26, 28, 30
1. Admission Charges for Movies The average admission charge for a movie is $5.39. If the
distribution of admission charges is normal with a standard deviation of $0.79, what is the probability that
a randomly selected admission charge is less than $3.00?
z
X   $3.00  $5.39 $2.39


 3.04

$0.79
$0.79
then look up z = -3.04 in Table E and you get that the area = 0.4988
P ( z  3.04)  0.5  0.4988  0.0012
-3.04
0
2. Teachers’ Salaries The average salary for first-year teachers is $27,989. If the distribution is
approximately normal with   $3250 , what is the probability that a randomly selected first-year
teacher makes these salaries?
a) Between $20,000 and $30,000 a year
z
X   $20, 000  $27,989 $7,989


 2.46

$3250
$3250
then look up z = -2.46 in Table E and you get that the area = 0.4931
z
X   $30,000  $27, 989 $2, 011


 0.62

$3250
$3250
then look up z = 0.62 in Table E and you get that the area = 0.2324
P (2.46  z  0.62)  0.4931  0.2324  0.7255
-2.46
0
0.62
b) Less than $20,000 a year
z
X   $20, 000  $27,989 $7,989


 2.46

$3250
$3250
P ( z  2.46)  0.5  0.4931  0.0069
-2.46
0
9. Microwave Oven Lines A survey found that people keep their microwave ovens an average of 3.2
years. The standard deviation is 0.56 year. If a person decides to buy a new microwave oven, find the
probability that he or she owned the old oven for the following amount of time. Assume the variable is
normally distributed.
a) Less than 1.5 years
z
X   1.5  3.2 1.7


 3.04

0.56
0.56
then look up z = -3.04 in Table E and you get that the area = 0.4988
P ( z  3.04)  0.5  0.4988  0.0012
-3.03
0
b) Between 2 and 3 years
z
X   2  3.2 1.2


 2.14

0.56
0.56
then look up z = -2.14 in Table E and you get that the area = 0.4838
z
X   3  3.2 0.2


 0.36

0.56
0.56
then look up z = 0.36 in Table E and you get that the area = 0.1406
P (2.14  z  0.36)  0.4838  0.1406  0.3432
-2.14
-0.36
c) More than 3.2 years
z
X   3  3.2
0


0

0.56
0.56
P ( z  0)  0.5
0
d) What percent of microwave ovens would be replaced if a warranty of 18 months were given?
For an 18 month (1.5 year) warranty,
z
X   1.5  3.2 1.7


 3.04

0.56
0.56
then look up z = -3.04 in Table E and you get that the area = 0.4988
P ( z  3.04)  0.5  0.4988  0.0012 Hence, 12% of ovens would be replaced.
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
z
X   1000  3262 2262


 2.06

1100
1100
hen look up z = -2.06 in Table E and you get that the area = 0.4803
P ( z  2.06)  0.5  0.4803  0.9803
-2.06
0
b) That the senior owes more than $4000
z
X   4000  3262 738


 0.67

1100
1100
hen look up z = 0.67 in Table E and you get that the area = 0.2486
P ( z  0.67)  0.5  0.2486  0.2514
0
0.67
c)
That the senior owes between $3000 and $4000
z
X   3000  3262 262


 0.24

1100
1100
then look up z = -0.24 in Table E and you get that the area = 0.0948
z
X   4000  3262 738


 0.67

1100
1100
then look up z = 0.67 in Table E and you get that the area = 0.2486
P (0.24  z  0.67)  0.0948  0.2486  0.3434
-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
z
X   15  23.5 8.5


 2.36

3.6
3.6
then look up z = -2.36 in Table E and you get that the area = 0.4909
z
X   22  23.5 1.5


 0.42

3.6
3.6
then look up z = -0.42 in Table E and you get that the area = 0.1628
P (2.36  z  0.42)  0.4909  0.1628  0.3281
-2.36
-0.42
b) Less than 18 minutes or more than 25 minutes
z
X   18  23.5 5.5


 1.53

3.6
3.6
then look up z = -1.53 in Table E and you get that the area = 0.4370
z
X   25  23.5 1.5


 0.42

3.6
3.6
then look up z = 0.42 in Table E and you get that the area = 0.1628
P ( z  1.53 or z  0.42)  (0.5  0.4370)  (0.5  0.1628)  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?
z
X   15  23.5 8.5


 2.36

3.6
3.6
then look up z = -2.36 in Table E and you get that the area = 0.4909
P ( z  2.36)  0.5  0.4909  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.
z
X   1200  949 251


 2.51

100
100
then look up z = 2.51 in Table E and you get that the area = 0.4940
P ( z  2.51)  0.5  0.4940  0.006
2.51
For the least expensive 10%, the area is 0.4 on the left side of the curve. Thus,
z value whose area is 0.4000 is z  1.28
x  1.28(100)  949  $821
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
1  0.05  0.95
0.95  0.5  0.45
In table E , find the z value that
corresponds to an area of 0.4500
z  1.64
x  z *  
x  1.64 *18  122.6  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
0.6 / 2  0.3
z value whose area is 0.3 is z  0.84, 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
0.97  0.5  0.47
z value whose area is 0.47 is z  1.88
Min  z *     1.88*100  400  588
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
1  0.015  0.985
0.985  0.5  0.485
z value whose area is 0.485 is z  2.17
Min  z *     2.17 *100  400  183
28. Cost of Groceries If a one-person household spends an average of $40 per week on groceries; find
the maximum and minimum dollar amounts spent per week for the middle 50% of one-person
households. Assume that the standard deviation is $5 and the variable is normally distributed.
  $40,  $5
0.5 / 2  0.25
z value whose area is 0.2500 is z  0.66,0.66
Min  z *     0.66*5  40  $36.7
Max  z *     0.66*5  40  $43.3
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
0.85  0.5  0.35
z value whose area is 0.3500 is z  1.04
Cutoff  z *     1.04*8  62  70.32
6.5 (p.325) #9, 13, 15, 17, 21, 23
9. Cost of Dog Ownership The average yearly cost per household of owning a dog is $186.80. Suppose
that we randomly select 50 households that own a dog. What is the probability that the sample mean
for these 50 households is less than $175.00? Assume σ= $32.
µ=186.80, n=50, σ= 32, P( x < 175.00)
Z
x   x   175.00  186.80 175.00  186.80 11.80



= 2.61

32
32


4.526
7.07
50
n
P ( z  2.61)  0.4955
Since we want the area to the left of z=-2.61, subtract 0.4955 from 0.5000
=0.0045 or 0.45%
0
12. Teachers’ Salaries The average teacher’s salary in North Dakota is $29,863. Assume a normal
distribution with α =$5100.
a. What is the probability that a randomly selected teacher’s salary is greater than $40,000?
x   40,000  29,863

 1.99

5100
P( z  1.99)  .4767
P( z  1.99)  0.5  0.4767  0.0233
z
The probability that a randomly selected teacher’s salary is greater than $40,000 is 2.33%
b. What is the probability that the mean for a sample of 80 teachers’ salaries is greater than $30,000?
x   30,000  29,863

 0.24

5100
P( z  0.24)  0.0948
P( z  0.24)  0.5  0.0948  0.4052
z
The probability that the mean for a sample of 80 teachers’ salaries is greater than $30,000 is about 40%.
13. Price of Bacon The average price of a pound of sliced bacon is $2.02. Assume the standard
deviation is $0.08. If a random sample of 40 one-pound packages is selected, find the probability that
the mean of the sample will be less than $2.00.
µ=2.02, n=40, σ= 0.08, P( x < 2.0)
Z
0.02
2.00  2.02 0.02


 1.58
0.08
0.08
0.01265
6.324
40
P ( z  1.58)  0.4429
Therefore, the P( x < 2.0)= 0.5 – 0.4429= 0.0571 or 5.71%
15. Sodium in Frozen Food The average number of milligrams (mg) of sodium in a certain brand of lowsalt 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)
Z
670  660 10

 0.285  0.29
35
35
P ( z  0.29)  0.1141
P( x < 670) = 0. 5 – 0.1141 = 0.3859 or 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)
Z
10
10
670  600


 0.902  0.90
35
35
11.08
3.16
10
P ( z  0.90)  0.3159
P( x <670) = 0.5 – 0.3159 = 0.1841 or 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)
Z
3
3
126  123


 0.55
21
21
5.43
3.87
15
P ( z  0.55)  0.2088
By symmetry, 0.2088 x 2 = 0.4176 or 41.76%
120
123 126
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( x <43) normally distributed
Z
43  46.2 3.2

 0.4
8
8
P ( z  0.4)  0.1554
P( x <43) = 0.5 – 1.554 = 0.3446 or 34.46%
b. Find the probability that, if 50 randomly selected adults take the test, the mean time it takes the
0.4
group to
complete the test will be less than 43 minutes.
µ=46.2, σ= 8, n=50 P( x <43)
z
43  46.2 3.2 3.2


 2.83
8
8
1.13
7.07
50
P ( z  2.83) = 0.4977
P( x <43)=0.5 – 0.4977= 0.0023 or 0.23%
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
Z
220  215 5

 0.33
15
15
P ( z  0.33) = 0.1293
P( x >220) = 0.5 – 0.1293= 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)
z
5 5
220  215

  1.67
15 3
15
5
25
P ( z  1.67) = 0.4525
P( x >220) = 0.5 – 0.4525 = 0.0475 or 4.75%
Ch 6.6 p.333 #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. College Enrollment The percentage of Americans 25 years or older who have at least some college
education is 50.9%. In a random sample of 300 Americans 25 years old and older, what is the probability
that more than 175 have at least some college education?
n=300, p=0.509
q = 1–q = 1 – 0.509 = 0.491
Since np = (300)(0.509) = 152.7 and nq = (300)(0.491) = 147.3, the normal distribution can be used to
approximate the binomial distribution.
µ = np = (300)(0.509) = 152.7
σ=
npq =
(300)(0.509)(0.491) =
74.9757 = 8.65885
For binomial distribution, P(X >175)
For normal distribution, P(X >175 + 0.5) = P(X > 175.5)
0.4957
152.7
The z value is z =
175.5
175.5  152.7
= 2.63
8.65885
From table E (Normal Distribution ), z=2.63 → 0.4957
0.5 – 0.4957 = 0.0043 or 0.43%
8. Personal Computers According to recent surveys, 53% of households have personal computers. If a
random sample of 175 households is selected, what is the probability that more than 75 but fewer than
110 have a personal computer?
n=175, p=0.53
q = 1–q = 1 – 0.53 = 0.47
Since np = (175)(0.53) = 92.75 and nq = (175)(0.47) = 82.25, the normal distribution can be used to
approximate the binomial distribution.
µ = np = (175)(0.53) = 92.75
σ=
npq =
(175)(0.53)(0.47) =
43.5925 = 6.60246
For binomial distribution, P(X > 75) and P(X < 110)
For normal distribution, P(75 + 0.5 < X < 110 – 0.5)
= P(75.5 < X < 109.5)
0.4945
75.5
z1 =
0.4955
92.75
109.5
109.5  92.75
= 2.54
6.60246
z2 =
75.5  92.75
= –2.61
6.60246
From table E (Normal Distribution), z1=2.54 → 0.4945, and z2=2.61 → 0.4955
0.4955 + 0.4945 = 0.99 or 0.99%
11. Female Scientists and Engineers Women comprise 83.3% of all elementary school teachers. In a
random sample of 300 elementary school teachers, what is the probability that more than 50 are men?
n=300, p=0.833
q = 1–q = 1 – 0.833 = 0.167
Since np = (300)(0.833) = 249.9 and nq = (300)(0.167) = 50.1, the normal distribution can be used to
approximate the binomial distribution.
µ = np = (300)(0.833) = 249.9
σ=
npq =
(300)(0.833)(0.167) =
41.7333 = 6.4601
When P(X > 50) for women for binomial, this means P(X < 250) for men.
For normal distribution, P(X < 250 – 0.5)
= P(X < 249.5)
0.4761
249.5
z=
249.9
249.5  249.9
= –0.06
6.4601
From table E (Normal Distribution), z=0.06 → 0.0239
0.5 –0.0239 = 0.4761 or 47.61%