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Welcome to Pstat5E:
Statistics with Economics and
Business Applications
Solution to Practice Midterm Exam
Yuedong Wang
Practice Midterm
1. The probability distribution of a discrete
random variable X is shown in the following
table:
x
1
P(X=x) .2
2
.3
3
p
4
.1
5
.2
(a) Find the missing value p.
(b) Find P(X<2).
(c) Find mean and variance of the random
variable X.
Practice Midterm
Solution:
(a) .2+.3+p+.1+.2=1  p=.2
(b) P(X<2) = P(X=1) = .2
(c) mean
= 1 x .2+ 2 x .3 + 3 x .2 + 4 x .1 + 5 x .2
= 2.8
variance
= (1-2.8)2 x .2 + (2-2.8)2 x .3 + (3-2.8)2 x .2 +
(4-2.8)2 x .1 + (5-2.8)2 x .2
= 1.96
Practice Midterm
2. The following are ages of 5 UCSB students, randomly
selected from those who have lunch in the UCen:
21, 18, 19, 31, 22
(i) Find mean, median and inter-quartile range.
(ii) Construct box plot.
(iii) What percentage of the students are teenagers?
Practice Midterm
Solution:
(i)
mean=(21+18+19+31+22)/5=22.2
sort observations: 18, 19, 21, 22, 31
.5(n+1)=3  median=21
.25(n+1)=1.5  Q1=(18+19)/2=18.5
.75(n+1)=4.5  Q3=(22+31)/2=26.5
IQR=Q3-Q1=26.5-18.5=8
Practice Midterm
(ii) Lower fence=18.5-1.5*8=6.5
Upper fence=26.5+1.5*8=38.5
18
18.5
21
26.5
31
(iii) Percentage of teenagers=2/5=40%
Practice Midterm
3. When I play soccer, one outcome of interest is the result of the
game. The possibilities are that my team wins, loses, or ties.
Another outcome of interest is whether or not I score (at least
one goal). You are given the following probabilities: the
probability that I score and my team wins is .3; the probability
that I score and my team loses is .15; the probability that my
team wins is .4; the probability that my team loses is .5; the
probability that I score is .5.
(i) Given that my team wins, what is the probability that I score?
(ii) Given that I score, what is the probability that my team wins?
(iii) What is the probability that my team loses or I do not score?
(iv) What is the probability that I do not score and my team ties?
Practice Midterm
Solution: Define events
W: my team wins, L: my team loses, T: my team ties
S: I score, N: I do not score. Then
P(W)=.4, P(L)=.5, P(S)=.5, P(SW)=.3, P(SL)=.15
 P(N)=1-P(S)=.5, P(T)=1-P(W)-P(L)=.1
(i)
P(S|W) = P(SW)/P(W) = .3/.4 = .75
(ii) P(W|S) = P(SW)/P(S) = .3/.5 = .6
Practice Midterm
Solution:
(iii) P(L)  P(LS)  P(LN) 
P(LN)  P(L) - P(LS)  .5 - .15  .35
P(L  N )  P(L)  P(N) - P(LN)  .5  .5 - .35  .65
(iv) P(T)  P(ST)  P(NT)  P(NT)  P(T) - P(ST)
P(S)  P(SW)  P(SL)  P(ST)

P(ST)  P(S) - P(SW) - P(SL)  .5 - .3 - .15  .05
P(NT)  P(T) - P(ST)  .1 - .05  .05
Practice Midterm
4. The menu at the Coffee Garden at 900 East and 900 South in
Salt Lake City has included a scrumptious selection of
quiche for about 10 years. The recipe calls for four fresh
eggs for each quiche. A Salt Lake County Health Department
Inspector paid a visit recently and pointed out that research
by the Food and Drug Administration indicates that one in
four eggs carries salmonella bacterium, so restaurants should
never use more than three eggs when preparing quiche. The
manager on duty wondered aloud if simply throwing out
three eggs from each dozen and using the remaining nine in
four-egg-quiches would serve the same purpose. The
inspector wasn't sure, but she said she would research it.
Salt Lake Tribune, 11 October 2002.
Practice Midterm
(a) What is the probability that
(i) of 4 randomly selected eggs, at least one carries
salmonella bacterium?
(ii) of 3 randomly selected eggs, at least one carries
salmonella bacterium?
(b) How many eggs would you expect to carry
salmonella bacterium our of
(i) 3 randomly selected eggs?
(ii) 4 randomly selected eggs?
Practice Midterm
Solution:
(a) (i) success  carries salmonella bacterium,
n  4, p  .25, q  .75,
P(x1)  1-P(x 0)  1- .250 .754  .684
(ii) n 3, p .25, q .75,
P(x1)  1-P(x 0)  1- .753  .578
(b) (i) n  4, p .25,
(ii) n 3, p .25,
  np  4.25 1
  np  3.25 .75
Practice Midterm
5. Tom and Dick took the same class. The final
scores for the class were normally distributed
with mean 75 and standard deviation 10.
Tom's score was 60 and Dick's score is better
than 80%, but worse than 20% of all scores.
(i) What is Dick's score?
(ii) What proportion of scores is better than
Tom’s?
Practice Midterm
Solution:
Final scores have a normal distributi on with   75 and   10.
(i) Denote Dick's score as "?". Then
x -75 ?75
?75
?75
.8  P(x?)  P(

)  P(z
) .3 P(0 z
).
10
10
10
10
?75
From Table 3,
.84  ?7510 .84  83.4
10
x -75 6075
(ii) P(x 60)  P(

)  P(z -1.5)
10
10
 P(-1.5 z0)  P(z0)  P(0 z 1.5) .5  .4332  .5  .9332
Practice Midterm