Download Exercise for Z table

Exercise (Chapters 7 & 8):
Chapter 7
1. The average weekly starting salary of computer programmers in a city is $750 with a standard
deviation of $40. Assume that we select a random sample of 64 computer programmers.
(a) Show the sampling distribution of the sample means.
(b) What is the probability that the sample mean will be greater than $740?
Ans.
(a)  x    750 ,
x 

n

40
64
 5,
The sampling distribution of X is a normal distribution N(750, 5).
(b) P( X  740)  .5  .4772  .9772
The following procedure is very helpful:
z
740  750
 2
5
Z Table: for z = -2, we have .4772
740
750
z
0
2. There are 8,000 students at the University of Tennessee at Chattanooga. The average age of all
the students is 24 years with a standard deviation of 9 years. A random sample of 36 students
is selected.
(a) Determine the standard error of the mean.
(b) What is the probability that the sample mean will be larger than 25.5?
(c) What is the probability that the sample mean will be between 21.6 and 27 years?
Ans.
(a)  x 
9
 1.5
36
(b) P( X  25.5)  .5  .3413  .1587
The following procedure is very helpful:
25.5  24
1
1.5
z
24
25.5
0
z
21.6
24
27
z1
0
z2
Z Table: for z = 1, we have .3413
(c) P(21.6  X  27)  .4452  .4772  .9224
The following procedure is very helpful:
z1 
21.6  24
 1.6
1.5
27  24
z2 
2
1.5
Z Table: for z1 = -1.6, we have .4452,
for z2 = 2, we have .4772
3. A department store has determined that 25% of all their sales are credit sales. A random
sample of 60 sales is selected.
(a) What is the sampling distribution of sample proportions?
(b) What is the probability that the sample proportion will be greater than 0.30?
(c) What is the probability that the sample proportion will be between 0.20 and 0.30?
Ans.
(a)  p  p  .25 ,
p 
p(1  p)
.25(1  .25)

 .0559 ,
n
60
The sampling distribution of p is approximately a normal distribution N(.25, .0559).
(b) P( p  .3)  .5  .3133  .1867
The following procedure is very helpful:
z
.25
.3
0
z
.3  .25
 .89
.0559
Z Table: for z = 1, we have .3133
(c) P(.2  p  .3)  .3133  .3133  .6266
The following procedure is very helpful:
z1 
.2  .25
 .89
.0559
z2 
.3  .25
 .89
.0559
.2
.25
.3
z1
0
z2
Z Table: for z1 = -.89, we have .3133,
for z2 = .89, we have .3133
Chapter 8
4. A random sample of 81 credit sales in a department store showed an average sale of $68.00.
From past data, it is known that the standard deviation of the population is $27.00.
(a) Determine the standard error of the mean.
(b) At 95% confidence, what is the margin of error?
(c) What is the 95% confidence interval of the population mean?
Ans.
(a)  x 

n

27
81
3
(b) It’s a large-sample case for estimation of population mean – use Z table. Because 1-α = .95,
probability = (1-α)/2 = .475. In the Z table, the Z value corresponding to probability = .475 is
1.96. Therefore,

 1.96
27
 5.88
n
81
(c) UCL = X  Margin of error  68  5.88  73.88
LCL = X  Margin of error  68  5.88  62.12
Margin of error = Z
5. The onwer of a gift store in Rock Hill wanted to determine the average age of his customers. A
random sample of 25 customers revealed an average age of 32 years with a standard deviation
of 8 years. Determine a 95% confidence interval estimate for the average age of all his
customers. Assume the population ages has a normal distribution.
Ans.
It’s a small-sample case for estimation of population mean as the population standard deviation is
unknown – use t table. Because 1-α = .95, upper-tail probability = α/2 = 0.025. And, degrees of
freedom = n-1 = 25-1 = 24. In the t table, the t value corresponding to upper-tail probability =
.025 and d.f. = 24 is 2.064. Therefore,
s
 2.064
8
 3.3024
n
25
UCL = X  Margin of error  32  3.3024  35.3024
LCL = X  Margin of error  32  3.3024  28.6976
Margin of error = t
6. A local health care facility noted that in a sample of 200 patients, 180 were referred to them by
the local hospital. Provide a 99% confidence interval for the proportion of all the patients who
were referred to this facility by the local hospital.
Ans.
It’s a large-sample case for estimation of population proportion – use Z table. Because 1-α = .99,
probability = (1-α)/2 = 0.495. In the Z table, the Z value corresponding probability = .495 is 2.57.
Therefore,
180
 .9
200
p(1  p)
.9(1  .9)
Margin of error = Z
 2.57
 .0545
n
200
UCL = p  Margin of error  .9  .0545  .9545
LCL = p  Margin of error  .9  .0545  .8455
Sample proportion p 