ASSIGNMENT 06
MA260 Statistical Analysis I
Directions: Be sure to make an electronic copy of your answer before submitting.
Unless otherwise stated, answer in complete sentences, and be sure to use correct English
spelling and grammar. Sources must be cited in APA format. Your response should be a
minimum of one (1) single-spaced page to a maximum of two (2) pages in length; refer to
the "Assignment Format" page for specific format requirements.
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.
75  90
 0.682
22
100  90
z2 
 0.455
22
z1 
The labor cost for the first day was 0.682 standard deviations below the mean and the
labor cost for the second day was 0.455 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?
z
415  400
 1.5
10
Area = 0.4332
b. What is the area between the mean and 395 pounds?
z
395  400
 0.5
10
Area = 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?
P( x  395)  P( z  0.5)  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?
P(x > A) = 0.05
z = 1.6449
A  1200
 1.6449  A  1570
225
4. Research on new juvenile delinquents revealed that 38% of them committed another
crime.
Normal approximation:
Mean = np = 100(0.38) = 38
Standard deviation = np(1  p)  100(0.38)(1  0.38)  4.854
a. What is the probability that of the last 100 new juvenile delinquents put on
probation, 30 or more will commit another crime?
29.5  38 

P( x  30)  P z 
  P( z  1.751)  0.96
4.854 

b. What is the probability that 40 or fewer of the delinquents will commit another
crime?
40.5  38 

P( x  40)  P z 
  P( z  0.515)  0.6967
4.854 

c. What is the probability that between 30 and 40 of the delinquents will commit
another crime?
40.5  38 
 29.5  38
P(30  x  40)  P
z
  P(1.751  z  0.515)  0.6567
4.854 
 4.854
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?
25 x
1
 0.71
35
b. Use the normal approximation to estimate the probability of at least one
disaster in 25 missions.
Mean = np = 25(1/35) = 0.71
Standard deviation = np(1  p)  25(0.71)(1  0.71)  0.833
0.5  0.71 

P( x  1)  P z 
  P( z  )  0.252  0.5995
0.833 

This is the end of Assignment 06.