Exam 2
DS-203
Fall 2004
Name___________________
Show all your work.
1.
In a sodium dietary study, the initial systolic blood pressure (in mm Hg) for all
female participants is found to have a mean value of 120.5 mm and a standard
deviation of 12.4 mm. Let x denote the initial systolic blood pressure for a female
participant in this study.
a.
If the distribution of x is normal and a sample of n = 16 female subjects is
randomly selected, what is the probability that the sample mean x is
between 115.5 mm and 123.5 mm?
P (11.5 < x < 123.5) = P (115.5-120.5 < 123.5-120.5)
12.4 / √16
b.
12.4 / √16
= P (-1.61 < z < .97) = .8340 - .0537 = .78
What is the probability that the sample mean is at most 125.5 mm?
P ( x ≤ 125.5 ) = P( z < 125.5 -120.5) = P ( z < 1.61)
12.4 / √16
= .9463
c.
What is the initial blood pressure (in mmHg) of the upper 5% of all the
female participants?
.95
5%
x ___________________________
120.5
5%
______________________ z
0
From z-table
Z .95 = 1.645
Z = x-µ

1.645 = x - 120.5
12.4
x = (1.645)( 12.4) + 120.5
x = 140.898
2.
You are planning a survey of starting salaries for recent business major graduates
from your college. From a pilot study you estimate that the standard deviation is
about $8000. What sample size do you need to have a margin of error equal to
$500 with 95% confidence?
 = $ 8000
Margin of error = Z* __  __
n
Z * For 95% Confidence = 1.96
8000
n
500 = 1.96
500 n = 1.96 (8000)
n = 15680 = 31.36
500
n = 983.45  984
3.
You want to rent an unfurnished one-bedroom apartment for next semester.
The mean monthly rent for a random sample of 100 apartments advertised in
the local newspaper is $540. Assume that the standard deviation is $80. Find a
98% confidence interval for the mean monthly rent for unfurnished onebedroom apartments available for rent in this community.
n = 100
x = 540
 = 80
Z* for 98% confidence interval = 2.33
x  Z* __  __
n
__80__
540  2.33 100
540  (2.33) (8)
540  18.64
(521.36
558.64)
4.
A large manufacturing plant has averaged 7 “reported accidents’ per month.
Suppose that accident counts over time follow a poison distribution with mean
7 per month.
a) What is the probability of 7 accidents in a month?
P( x = k) = e   µ 
k!
P( x = 7) = e 7 7 7 = (0.0009119)(823543)
7!
5040
= 750.974 = .149
5040
b) What is the probability of at leas 2 accidents in a month?
P ( x  2) = 1- [ P (x = 0) + P (x = 1)]
= 1 – [e 7 7 0 + e 7 7 1 ]
0!
1!
5.
= 1 – (.0009119 + .0063832)
= 1 - .0072951 = .9927
You have two scales for measuring weight. Both scales give answers that vary
a bit in repeated weightings of the same item. If the true weight of an item is 2
grams (g), the first scale produces readings X that have mean 2.000 g and
standard deviation .002 g. The second scale’s readings Y have mean 2.001g
and standard deviation .001 g.
a) What are the mean and standard deviation of the difference Y-X between
the readings? (The readings X and Y are independent.)
 y x =  y -  x = 2.001-2 = .001

2
y x

=
=
y x

2
y
+ 
2
x
= (.001) 2 + (.002) 2 = .000005
.000005 = .0022
b) You measure once with each scale and average the readings. Your result is
Z = (X+Y)/2. What are z and z?
z =

1
1
 x +  y = ( 2 + 2.001) = 2.0005
2
2
2
z

z
1 2 2
1
)  x y = ( ) 2 ( .0022)
2
2
1
= ( .0022) = .0011
2
=(
2
c) Is the average Z more or less variable than the reading Y of the less
variable scale?
More variable .0011 vs. .001
Multiple Choice Questions
Select the best answer.
1.
A sample of size 5 is to be selected from a population whose distribution is
normal.
a.
The x sampling distribution will be normal.
A
b.
We can’t assume that thex sampling distribution will be normal.
c.
None of the above.
2.
A sample of size 50 is to be selected from a population whose distribution is
positively skewed.
a.
The x sampling distribution will be approximately normal.
A
b.
We can’t assume that thex sampling distribution will be normal.
c.
None of the above.
3.
For which of the following sample sizes will the standard deviation of the x
distribution be the smallest?
a.
n=5
C
b.
n = 10
c.
n = 20
d.
It will be the same for all.
4.
If a population has a standard deviation , then the standard deviation of mean of
100 randomly selected items from this population is
(a) 
C
(b) 100
(c) /10
(d) /100
(e) 0.1
Use the following problem for questions 5-7.
Let X denotes the number of siblings for a college professor at a particular university.
Suppose that X has the following probability distribution.
X
0
1
2
3
4
5
6
P(X)
.2
.3
.15
.13
.12
.08 .02
5.
6.
What proportion of professors at this university are only children?
a) .5
b) .2
c) .65
d) .73
e) None of these
B
What is the probability that a randomly selected professor has two or fewer
siblings?
a) .5
b) .2
c) .65
d) .73
e) None of these
C
7.
What is the probability that a randomly selected professor comes from a family
with 2 or fewer children?
a) .5
b) .2
c) .65
d) .73
e) None of these A
Use the following problem for questions 8 and 9.
Parking violations in a particular city result in a fine of either $5, $10, or $25, depending
on the type of violation. Suppose that X = amount of fine has the following probability
distribution.
X
5
10
25
P(X)
.2
.70
.10
8.
The mean value of X is.
B
a) 5.22
b) 10.50
c) 27.25
d) 6300
e) None of these
9.
The standard deviation of X is.
a) 5.22
b) 10.50
c) 27.25
d) 6300
e) None of these A
10.
11.
12.
x =
x =
Which of the following variables is not discrete?
a) The number of students absent from a class.
D
b) The number in a group of 20 people who have college degrees.
c) The number of students in a class who earned an A on the midterm exam.
d) The dept of a crack in dry clay.
e) None of the above.
A random sample is to be selected from a population with mean  = 100 and
standard deviation  = 20. Determine the mean of the x distribution for a sample
of size 16.
a.
 = 100
A
b.
 = 100/16
c.
 = 100/4
d.
None of the above
A random sample is to be selected from a population with mean  = 100 and
standard deviation  = 20. Determine the standard deviation of the x distribution
for a sample of size 16.
a.
 = 20
C
b.
 = 20/16
c.
 =20/4
d.
None of the above
 x P (x) = 5 (.2) + 10 (.7) + 25 (.1) = 10.5
x   x  P( x) =
2
=
5  10.52 (.2)  (10  10.5) 2 (.7)  (25  10.5) 2 (.1)
27.25 = 5.22