CHAPTER 2: SAMPLE PROBLEMS FOR HOMEWORK, CLASS OR EXAMS
These problems are designed to be done without access to a computer, but they may require
a calculator.
1. a. You have collected data on debt ratios (debt divided by annual income) among
households in your county. The distribution is very positively (right) skewed. The most
appropriate probability distribution for modeling the individual observations in this data
would be:
___ normal distribution
___ binomial distribution
2
___  distribution
___ Poisson distribution
b. You have collected data on numbers of power outages each month at your home. The most
appropriate probability distribution for modeling this data would be:
___ normal distribution
___ binomial distribution
2
___  distribution
___ Poisson distribution
c. You have collected data on debt ratios (debt divided by annual income) among 500
randomly selected households in your county, and computed the resulting sample mean. The
most appropriate probability distribution for modeling the sample mean would be:
___ normal distribution
___ binomial distribution
2
___  distribution
___ Poisson distribution
d. Each month, you randomly survey a sample of 200 adults in your metropolitan area to ask
them whether or not they are ‘confident’ in their financial security. The most appropriate
probability distribution for modeling the number who answer Yes would be:
___ normal distribution
___ binomial distribution
2
___  distribution
___ Poisson distribution
2. The temperature readings in a thermocouple fluctuate with a normal distribution that has a
mean of 800 C and a standard deviation of 4 C.
a. What is the probability the temperature will be between 798 and 807 C ?
b. Give a value that separates the lowest 20% of the temperatures from the upper 80%.
Draw the sketch that accompanies this question.
3. Even when a manufacturing line is working properly, each resistor it produces has a small
(4%) chance of being defective. At the end of each shift, the supervisor randomly selects 20
independent resistors and tests them. If two or more of the test sample is defective, then
the manufacturing line is halted for recalibration.
a. What is the probability the line will be halted, even though the probability of being
defective is really its usual 4%?
b.Suppose the line has slipped out of adjustment, and the probability for each resistor of
being defective is now 10%. What is the probability the supervisor will not detect the
problem, and the line will not be halted for calibration?
4. Specifications for a monofilament line state that 95% of the specimens must have a
diameter between 3.5 and 4.0 mm.. The diameters follow a normal distribution, and you
have adjusted the manufacturing machine so the mean  = 3.75 mm. What must  be?
Begin by showing the appropriate sketch for this problem.
5. The Department of Environmental Protection demands that your power plant show a
mean NOx emission less than 5 lb/MWh. Your plant’s daily NOx emission actually
fluctuates randomly with a mean of 4.9 lb/MWh and a standard deviation of 0.5 lb/MWh.
If the DEP randomly selects and tests emissions on 30 independent days, what is the
probability that the sample mean will exceed the limit?
6. Your hospital tracks the monthly number of hospital-acquired infections. Past experience
shows this number follows a Poisson distribution with a mean of 4 infections per month.
a. What is the probability that a randomly selected month will have no hospital-acquired
infections?
b. Using the properties of the Poisson distribution, give the appropriate control limits for a
control chart for the number of infections.
7. An emergency management district has collected data on the number of people per
household that would need special assistance in the case of a prolonged power outage (X).
The probability distribution is shown below.
x
0
1
2
P(x) 0.98 0.015 0.005
a. Calculate the expected value and the variance for X.
b. A community has 10,000 households. Identify the distribution for the number of
households that have at least 1 person needing special assistance.
8. The sketch below shows a diagram where every component of the system is independent
of the others. Components labeled ‘A’ are expensive, and work with probability 0.99.
Components labeled ‘B’ are less expensive, and work with probability 0.95. Calculate the
reliability of the system.
B1
bB1
A1
B2
SOLUTIONS
1 a)  2 distribution
b) Poisson distribution
c) normal distribution
d) binomial distribution
2. a)
Your method will vary a little bit depending on whether you use the calculator or the paper
table. From the calculator normcdf(798,807,800,4) = .6514. From the paper table, Z = (798800)/4 = -.5 and for the upper limit Z = (807-800)/4 = 1.75, so probability = .959941 .308538 = .6514
b)Your sketch should show .2 of the probability to the left of the dividing line.
On the TI83, X = invnorm (.2,800,4) = 796.64,
Or using the table, Z = -.84, so X = 800 -0.84*4 = 796.64
3. a) X = number of defective in sample is binomial with n=20 and p = .04
Want P(X  2) = 1 – [ P(X=0)+P(X=1) ] = .1897
b) X = number of defective in sample is binomial with n = 20 and p = .10
Want P(X  1) = .3918
4. Sketch will have .95 of the probability between 3.5 and 4.0, and .025 of the probability in
each of the tails. Hence the lower limit of 3.5 must match a Z-score of -1.96, yielding the
equation
3.5  3.75
1.96 
   .1276

5. Want probability that x  5.0 We know that sample mean has  x  4.9 and standard
error  x  .5/ 30  .0913 . Since n is large, the sample mean is normally distributed. Using
calculator or normal probability table P( x  5)  P( Z  1.095)  .1367
minor variations on the answer will occur due to different ways people rounded or entered
the z-table.
6. a. e4  0.0183
b. The standard deviation would be 4  2 . Using the 3-sigma rule, the upper control limit
would be 4+3*2 = 10. The lower limit would be 4-3*2 = -2, but since the number can’t be
negative, the lower limit would effectively be 0.
7 a) expected value = 0.025, variance = 0.03438
b) binomial distribution with n = 10000 and p = 0.02.
8. P( [B1 or B2 work] and A1 works ) = 0.9975 * 0.99 = 0.9875