1. Compute the mean and variance of the following discrete probability distribution.
x
P(x)
2
.50
8
.30
10 .20
Work:
Mean = E(X)= 2(0.5)+8(0.3)+10(0.2)=1+2.4+2=5.4
V(X)=E(X^2)-E(X)^2
E(X^2)= 22(0.5)+82(0.3)+102(0.2)= 2+19.2+20=41.2
V(X)=41.2-5.42=12.04
Answer: mean is 5.4 and variance is 12.04
2. The Computer Systems Department has eight faculty, six of whom are tenured. Dr.
Vonder, the chair, wants to establish a committee of three department faculty members to
review the curriculum. If she selects the committee at random: a. What is the probability all
members of the committee are tenured? b. What is the probability that at least one member
is not tenured? (Hint: For this question, use the complement rule.)
Work:
a)
6C3/ 8C3 =
20/56=5/14
b) 1-6C3/ 8C3 = 1-5/14=9/14
Answer:
a) 5/14
b) 9/14
3. New Process, Inc., a large mail-order supplier of women’s fashions, advertises same-day
service on every order. Recently, the movement of orders has not gone as planned, and
there were a large number of complaints. Bud Owens, director of customer service, has
completely redone the method of order handling. The goal is to have fewer than five
unfilled orders on hand at the end of 95% of the working days. Frequent checks of the
unfilled orders follow a Poisson distribution with a mean of two orders. Has New Process,
Inc. lived up to its internal goal? Cite evidence.
Work:
Let X be the number of unfilled orders at the end of the day.
X has the Poisson distribution with parameter λ = 2
In general you have:
X ~ Poisson( λ )
P(X = x) = e-λ λx / x! for x = 0, 1, 2, 3, 4, ...
P(X = x) = 0 otherwise
In this problem we have
λ=2
X ~ Poisson( 2 )
P(X < 5 ) =
4
∑ P(X = i) = 0.947347
i=0
P(X = 0) = 0.13533528
P(X = 1) = 0.27067057
P(X = 2) = 0.27067057
P(X = 3) = 0.18044704
P(X = 4) = 0.09022352
Answer:
So no, they are close, but did not meet the internal goal. They will have less than five
unfilled orders 94.7347% of the time.
4. Recent information published by the U.S. Environmental Protection Agency indicates
that Honda is the manufacturer of four of the top nine vehicles in terms of fuel economy.
a. Determine the probability distribution for the number of Hondas in a sample of three cars
chosen from the top nine. b. What is the likelihood that in the sample of three at least one
Honda is included?
Work
X= number of Hondas in the sample
P(X=3) = 4C3/9C3 = 4/84 = 1/21
P(X=2) = (4C2)(5C1)/9C3 = 30/84 = 5/14
P(X=1) = (4C1)(5C2)/9C3 = 40/84 = 10/21
P(X=0) = 5C3/9C3 = 10/84 = 5/42
Answer:
x
P(x)
0
5/42
1
10/21
2
5/14
3
1/21
5.According to the “January theory,” if the stock market is up for the month of January, it
will be up for the year. If it is down in January, it will be down for the year. According to
an article in the Wall Street Journal, this theory held for 29 out of the last 34 years. Suppose
there is no truth to this theory. What is the probability this could occur by chance? NOTE:
Show your work in the problems.
Work:
There is no truth to this theory so we ca suppose that P(theory helds for a year)=0.5
X= number of years that theory helds
X has a binomial distribution with n=34 and p=0.5
P(X=29)= 34C29(0.5)29(0.5)5 =1.61966E-05
Answer: 1.61966E-05
6. A recent article in the Myrtle Beach Sun Times reported that the mean labor cost to
repair a color television is $90 with a standard deviation of $22. Monte’s TV Sales and
Service completed repairs on two sets this morning. The labor cost for the first was $75 and
it was $100 for the second. Compute z values for each and comment on your findings.
Work:
1) For x = 75 z = (75-90)/22 = -15/22
2) For x = 100 z = (100-90)/22 = 10/22 = 5/11
Answer: z-values are -15/22 and 5/11, the first sore is negative because 75<90 and the
second one positive because 100>90
7. The mean of a normal distribution is 400 pounds. The standard deviation is 10 pounds. a.
What is the area between 415 pounds and the mean? b. What is the area between the mean
and 395 pounds? c. What is the probability of selecting a value at random and discovering
that it has a value of less than 395 pounds?
Work:
a) P(400<X<415)=P(0<Z<15/10) = P(0<Z<1.5) = 0.433
b) P(395<X<400)=P(-5/10<Z<0)=P(-0.5<Z<0) = 0.191
c) P(X<395)=P(Z<-5/10)=P(Z<-0.5) = 0.309
Answer:
a) 0.433
b) 0.191
c) 0.309
8. The monthly sales of mufflers in the Richmond, VA area follow the normal distribution
with a mean of 1200 and a standard deviation of 225. The manufacturer would like to
establish inventory levels such that there is only a 5% chance of running out of stock.
Where should the manufacturer set the inventory levels?
Work:
X= monthly sales
Let the levels 1200-k and 1200+k so we have to find k where
P(1200-k<X<1200+k)=0.95
P(-k/225<Z<k/225) = 0.95
P(Z<k/225)=0.975
k/225 = 1.96
k=1.96*225=441
Levels are 1200-441=759 and 1200+441=1641
Answer: 759 and 1541
9. Research on new juvenile delinquents revealed that 38% of them committed another
crime. a. What is the probability that of the last 100 new juvenile delinquents put on
probation, 30 or more will commit another crime? b. What is the probability that 40 or
fewer of the delinquents will commit another crime? c. What is the probability that between
30 and 40 of the delinquents will commit another crime?
Work:
Let X= number of delinquents that commit another crime in the last 100 delinquents
X has a binomial distribution with n=100 and p=0.38
We have to find
a) P(X≥30)
b) P(X≤40)
c) P(30<X<40)
We can use normal approximation:
Y has a normal distribution with mean = np = 38
And sd = [np(1-p)] = [100*0.38*0.62]=4.854
a) P(X≥30) ≈ P(Y>29,5)=P(Z>(29,5-38)/4.854)=P(Z>-1,751)=0.96
b) P(X≤40) ≈ P(Y<40,5)=P(Z<(40,5-38)/4.854)=P(Z<0.515)=0.697
c) P(30<X<40) ≈ P(30.5<X<39.5)= P((30.5-38)/4.854<Z<(39.5—38)/4.854)
=P(-1.545<Z<0.309)=0.56
Answer:
a) 0.96
b) 0.697
c) 0.56
10. An Air Force study indicates that the probability of a disaster such as the January 28,
1986 explosion of the space shuttle Challenger was 1 in 35. The Challenger flight was the
25th mission. a. How many disasters would you expect in the first 25 flights? b. Use the
normal approximation to estimate the probability of at least one disaster in 25 missions.
Work:
X= number of disasters in 25 missions
X has a binomial distribution with n=25 and p=1/35
a) E(X)=np= 25/35=5/7
b) P(X≥1) = P(Y>0.5) where Y has a normal distribution with mean = 5/7
And sd = [np(1-p)] = [25*(5/7)*(2/7)]=5.102
P(Y>0.5)=P(Z>(0.5-5/7)/5.102)=P(Z>-0.042)=0.517
Answer:
a) 5/7
b) 0.517