ASSIGNMENT 06
MA260 Statistical Analysis I
NOTE: Show your work in the problems.
1. Compute the mean and variance of the following discrete probability distribution.
x
2
8
10
P(x)
.50
.30
.20
Mean = µ = å( X iP ( x )) = ( 2 ) ( 0.5 ) + ( 8 ) ( 0.3) + (10 ) ( 0.2 ) = 5.40
Variance = å éë X 2 iP ( x ) ùû - µ2
2
2
2
2
= éë( 2 ) ( 0.5 ) + ( 8 ) ( 0.3) + (10 ) ( 0.2 ) ùû - ( 5.40 )
= 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?
P ( all tenured ) =
C3 20 5
=
=
56 14
8 C3
6
b. What is the probability that at least one member is not tenured? (Hint: For this
question, use the complement rule.)
The probability that at least one is not tenured is the complement of the
probability that all three are tenured.
P ( at least one not tenured ) = 1- P ( all tenured ) = 1-
5
9
=
14 14
3. New Process, Inc., a large mail-order supplier of women’s fashions, advertises sameday 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.
The probability that the store has fewer than five unfilled orders is:
P ( X < 5, l = 2 ) = P ( 0,2 ) + P (1,2 ) + P ( 2,2 ) + P ( 3,2 ) + P ( 4,2 )
=
e-2 ( 2 0 )
0!
+
e-2 ( 21 )
1!
+
e-2 ( 2 2 )
2!
+
e-2 ( 2 3 )
3!
+
e-2 ( 2 4 )
4!
= 0.9473
On average, the supplier can expect to have fewer than five unfilled orders on
94.73% of the working days. This is technically less than the 95% goal, but is
extremely close. If the probability is rounded to the nearest whole percentage,
then the supplier can expect to have fewer than five unfilled orders on 95% of the
working days, thereby meeting their goal.
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.
C3 10 5
=
=
C
84
42
9 3
P (0) =
5
P (1) =
4
P (2) =
4
P ( 3) =
4
C1 i 5 C2 4 i10 40 10
=
=
=
84
84 21
9 C3
C2 i 5 C1 6 i5 30 5
=
=
=
84
84 14
9 C3
C3 4
1
=
=
84 21
9 C3
The distribution is then:
Number of cars
0
1
2
3
Probability
5/42
10/21
5/14
1/21
b. What is the likelihood that in the sample of three at least one Honda is included?
The probability that at least one Honda is include in the sample of 3 is:
P ( x ³ 1) = P ( x = 1) + P ( x = 2 ) + P ( x = 3)
=
10 5 1
+ +
21 14 21
=
37
42
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?
If there is no truth to the theory, then the probability that the year end result
matches with the January result would be p = 0.5.
Using the binomial probability calculation, the probability that the results would
match for 29 out of 40 years is:
P ( 29 out of 40 ) = C ( 40,29 ) * ( 0.5 ) * (1- 0.5 )
29
P ( 29 out of 40 ) = 0.0021
40-29
ASSIGNMENT 07
MA260 Statistical Analysis I
NOTE: Show your work in the problems.
1. 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.
For the $75 repair, the z value is:
z=
x- µ
s
=
75 - 90
= -0.6818
22
For the $100 repair, the z value is:
z=
100 - 90
= 0.4545
22
The z-values indicate that the $75 repair is 0.68 standard deviations below the
mean, while the $100 repair is 0.45 standard deviations above the mean.
2. 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?
Compute the z-value for 415:
z=
x- µ
s
=
415 - 400
= 1.5
10
The area between z = 1.5 and the mean (z = 0) is 0.4332.
b. What is the area between the mean and 395 pounds?
Compute the z-value for 395:
z=
x- µ
s
=
395 - 400
= -0.5
10
The area between z = -0.5 and the mean (z = 0) is 0.1915.
c. What is the probability of selecting a value at random and discovering that it has a
value of less than 395 pounds?
The probability is equal to the area below the z-value of -0.5.
The probability is 0.3085.
3. 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?
The inventory level needs to correspond with the point on the normal distribution
where the right tail area is 0.05. The z-value at this point is 1.645.
The corresponding inventory level is then:
x + zs = 1200 + (1.645 ) ( 225 ) = 1570.125
Since inventory level should be a discrete value, this should be rounded up to the
next highest integer.
The minimum inventory level should be 1571.
4. 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?
Using the normal approximation to the binomial distribution:
z = P ( x ³ 30 ) = P ( x > 29.5 )
The corresponding z value is then:
z=
x - np
=
npq
29.5 - ( 0.38 ) (100 )
(100 ) ( 0.38 )(1- 0.38 )
= -1.7512
Then
P ( x > 29.5 ) = P ( z > -1.7512 ) = 0.9600
b. What is the probability that 40 or fewer of the delinquents will commit another crime?
Using the normal approximation to the binomial distribution:
P(x ≤ 40) = P(x < 40.5)
The corresponding z value is then:
z=
x - np
=
npq
40.5 - ( 0.38 ) (100 )
(100 ) ( 0.38 )(1- 0.38 )
= 0.5151
Then
P ( x < 40.5 ) = P ( z < 0.5151) = 0.6968
c. What is the probability that between 30 and 40 of the delinquents will commit another
crime?
Using the normal approximation to the binomial distribution:
P(30 < x < 40) = P(30.5 < x < 39.5)
The corresponding z values are then:
z=
x - np
=
npq
z=
x - np
=
npq
30.5 - ( 0.38 ) (100 )
(100 ) ( 0.38 )(1- 0.38 )
39.5 - ( 0.38 ) (100 )
(100 ) ( 0.38 )(1- 0.38 )
= -1.5452
= 0.3090
Then
P ( 30.5 < x < 39.5 ) = P ( -1.5452 < z < 0.3090 )
= P ( z < 0.3090 ) - P ( z < -1.5452 )
= 0.6213 - 0.0611
= 0.5602
5. 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?
The probability of a disaster is p = 1/35.
The expected number of disasters in 25 missions is:
np = (25)(1/35) = 25/35 = 5/7
Expressed as a decimal, the expected number of disasters is 0.7143.
b. Use the normal approximation to estimate the probability of at least one disaster in 25
missions.
Using the normal approximation to the binomial distribution:
P(x ≥ 1) = P(x > 0.5)
The corresponding z-value is:
z=
x - np
np (1- p )
=
æ 1ö
0.5 - ( 25 ) ç ÷
è 35 ø
1
1
( 25 ) æçè ö÷ø æçè 1- ö÷ø
35
35
Then P(x > 0.5) = P(z > -0.2572) = 0.6015
= -0.2572