110
Math 116 – 02: Test #4 (Chapters 16, 18, 19)
Name:
Show work when appropriate to get full credit for correct answers.
Give grammatically correct explanations when asked to explain or support an answer.
1. Find the expected value (i.e. mean) of the random variable by hand.
computation not just an answer. (6 points)
x
100
200
300
400
P(x)
0.1
0.2
0.5
0.2

 
 
 

That is, show the
 
E x  100 0.1 200 0.2 300 0.5  400 0.2  280

2. The table below shows the probability model for the number of e-mails a certain college
professor receives from students during a day. Use your calculator to find the expected value
and standard deviation of the number of e-mails. (8 points)
# e-mails
Probability
0
1
2
3
4
5
0.05
0.10
0.20
0.25
0.30
0.10
1-Var Stats L1, L2


E x  2.95 e  mails,  x  1.32 e  mails
3. A dice game (played with one 10-sided die) costs $50 to play. On the first roll, if you roll a 2,
you receive $100 and the game is over. If not, then on the second roll if you roll a 5 or 10 you
receive $75 and the game is over. If not, then on the third roll if you roll an even number you
receive $35. Otherwise you receive nothing. Let x = the amount you profit in the game (so if
you win $100 on the first roll, you profit $50). Construct a probability model for x and give the
expected value and standard deviation of x. (14 points)
x
50
25





-15
-50
P(x)
1
10
= 0.1
   = 0.18
    = 0.36
    = 0.36
9
10
2
10
9
10
8
10
5
10
9
10
8
10
5
10


E x  $13.90,  x  $33.92
4. A report released in May 2005 by First Data Corp. indicated that 43% of adults had received a
“phishing” contact (a bogus email that replicates an authentic site for the purpose of stealing
personal information). Suppose a random sample of 400 adults is obtained.
)
(a) What is the mean and standard deviation of the p distribution? (8 points)
0.430.57
 p p  0.43 and  p
)
)
400

 0.0248

)
(b) Check the success/failure condition to ensure that the p distribution is approximately
normal. (4 points)
 
nq  4000.57 228
np  400 0.43  172


both are at least 10

(c) What is the probability that between 40% and 42% of the sample have received a
“phishing” contact? (6 points)




)
P 0.4  p  0.42  normalcdf 0.4, 0.42, 0.43, 0.0248  0.2302

(d) What is the probability that 45% or more of such a sample had received “phishing”
contacts? (6 points)




)
P p  0.45  normalcdf 0.45, "", 0.43, 0.0248  0.20999


(e) What is the smallest sample size that would satisfy the success/failure condition for the
)
p distribution to be approximately normal? (6 points)

We need both np and nq to be at least 10
 n 0.43  10 and n 0.57  10
 
 
 n  23.25 and n  17.5
So, n must be at least 24.

5. The Get-A-Grip tire company claims that, on average, the lifetime of its premium tires is 35000
miles with a standard deviation of 1750 miles. Suppose 100 of these tires are randomly
selected.
(a) What is the mean and standard deviation of the x distribution?
(8 points)
x    35000 miles and  x  1750  175 miles

100

(b) What is the probability that the average lifespan of this sample is between 34900 miles
and 35150 miles? (6 points)




P 34900  x  35150  normalcdf 34900, 35150, 35000, 175  0.5205

(c) What is the probability that the average lifespan of this sample of tires is less than
34000 miles? (6 points)




P x  34000  normalcdf "", 34000, 35000, 175  5.5E  9  0.0000000055

6. A sociologist wishes to estimate the percentage of Americans who favor affirmative action
programs for women and minorities for admission to colleges and universities. Assuming there
is no previous data to consider, what sample size should be obtained if she wishes to construct a
95% confidence interval with an error of no more than 3 percentage points? (8 points)
2
 z * 2
1.96 
n  0.25   0.25
  1067.111
0.03 
ME 
So, n must be at least 1068 people

7. A confidence interval is to be constructed for some population proportion.
(6 points each)
(a) Which will be wider, a 95% confidence interval or a 90% confidence interval? Explain.
A 95% C.I. is wider because having higher confidence in the answer requires a wider
range of possible values for the population proportion.
(b) For a 95% confidence interval, will a sample size of 100 or a sample size of 250 produce
a wider interval? Explain.
)
Sample size n = 100 is wider because the standard error for the p distribution, given
by
))
pq
n
, is bigger for smaller values of n.


8. In a Harris Poll conducted October 20 – 25, 2004, 2114 randomly selected adults who follow
professional football were polled, and 18% said that the Green Bay Packers were their favorite
team. (6 points each)
(a) Construct a 90% confidence interval for the proportion of all adults who follow
professional football who say the Green Bay Packers is their favorite team.




p  0.16648, 0.19398 if x = 381 is used
p  0.16602, 0.19349 if x = 380 is used


(b) Construct a 98% confidence interval for the proportion of all adults who follow
professional football who say the Green Bay Packers is their favorite team.




p  0.16078, 0.19968 if x = 381 is used
p  0.16033, 0.19918 if x = 380 is used

