Introduction to the Practice of Statistics
Sixth Edition
Moore, McCabe
Section 5.2 Homework Answers
5.43 A grinding machine for auto axles. An automatic grinding machine in an auto parts plant prepares
axles with a target diameter µ = 40.135 mm. The machine has some variability, so the standard deviation
of the diameters is σ = 0.003mm. A sample of 4 axles is inspected each hour for process control purposes,
and records are kept of the sample mean diameter. If the process mean is exactly equal to the target
value, what will be the mean and standard deviation of the numbers recorded?
σx =
µ x = 40.135mm
0.003
4
= 0.0015
5.45 Axle diameters. Averages are less variable than individual observations. Suppose that the axle
diameters in Exercise 5.43 vary according to a normal distribution. In that case, the mean x of an SRS of
axles also has a normal distribution.
a) Make a sketch of the normal curve for a single axle. Add the normal curve for the mean of an SRS of 4
axles on the same sketch.
40.126
40.129
40.132
40.135
40.138
40.141
40.144
b) What is the probability that the diameter of a single randomly chosen axle differs from the target value
by 0.006 mm or more?
Let the random variable X measure the diameter of an axle.
P( X < 40.129 OR X > 40.141) = 5% using the
68-95-99.7 Rule
40.129
40.135
40.141
c) What is the probability that the mean diameter of an SRS of 4 axles differs from the target value by
0.006mm or more?



40.129 − 40.135 
P( X < 40.129 OR X > 40.141) = 2 P  Z <

0.003


4


= 2P(Z < -4)
≈0
In actuality 0.0000634.
5.50 North Carolina State University posts the grade distributions for its courses online. You can find the
distribution grades in English 210in the Spring 2006 semester was
Grade
X
P(x)
A
4
0.31
B
3
0.40
C
2
0.20
D
1
0.04
F
0
0.05
a) Using the common scale A = 4, B = 3, C = 2, D = 1, F = 0, take X to be the grade of a randomly chosen
210 student. Use the definitions of the mean (page 271) and standard deviation (page 280) for discrete
random variables to find the mean µ and the standard deviation σ of grades in this course.
µX = 0.31(4) + 0.4(3) + 0.20(2) + 0.04(1) + 0.05(0) = 2.88
σX =
0.31(4 - 2.88) 2 + 0.4(3 - 2.88) 2 + 0.20(2 - 2.88) 2 + 0.04(1 - 2.88) 2 + 0.05(0 - 2.88) 2
σ = 1.051
b) English 210 is a large course. We can take the grades of an SRS of 50 students to be independent of
each other. If x is the average of these 50 grades, what are the mean and standard deviation of x ?
µ x = 2.88
while σ x ≈ 0.1486
c) What is the probability P(X ≥ 3) that a randomly chosen English 210 student gets a B or better? What
is the approximate probability P( x ≥ 3) that the grade point average for 50 randomly chosen English 210
students is B or better?
P(X ≥ 3) = 0.4 + 0.31 = 0.71



3-2.88 
P( x ≥ 3) ≈ P  Z >

1.051 



50 

≈ P(Z > 0.8074)
≈ 0.2097
5.52 A lottery payoff. A $1 bet in a state lottery's Pick 3 game pays $500 if the three-digit number you
choose exactly matches the winning number, which is drawn at random. Here is the distribution of the
payoff X:
Payoff X
$0
$500
Probability
0.999
0.001
Each day's drawing is independent of other drawings.
(a) What are the mean and standard deviation ofX?
µX = $0(0.999) + 500(0.001) = $0.5
σX =
0.999(0 − 0.5) 2 + 0.001(500 − 0.5) 2 = 15.80
(b) Joe buys a Pick 3 ticket twice week. What does the law of large numbers say about the average
payoff Joe receives from his bets?
I mentioned before that the sample space determines the probability. The problem I am having
here is that I can look at this question in two ways; I am not sure which one is the intent of the
question.
Case 1 – The mention of twice a week, is it to let me know that Joe is buying lots of tickets, but Iam
suppose to think of each ticket as an individual ticket. If that is the case the average payoff is $0.5
per ticket.
Case 2 - Is the mention of the two tickets to let me know that the sample space consists of the result
of two ticket outcomes, not just one. Then the average return per two tickets is 2(0.5) = $1. Thus, in
the long run, if you take years of buying 2 tickets a week, the average of those winnings would be
around $1. Of course this does not take into account that Joe on average exactly pays $2 per week
to have the privelage of getting $1 back on average.
(c) What does the central limit theorem say about the distribution of Joe’s average payoff after 104 bets in
a year?
The population of payoff is discrete, but it will have that roughly normal shape.
(d) Joe comes out ahead for the year if his average payoff is greater than $1. What is the probability Joe
ends the year ahead? I will think as a sample of 104. I am assuming from the previous discussion
that Joe buys 104 tickets in a year. I will think of the sample space as consisting of all single ticket
outcomes



1- (0.5) 
P( x ≥ 1) ≈ P  Z >

15.80 



104 

≈ P(Z > 0.3227)
≈ 0.3735
5.54. Flaws in carpets. The number of flaws per square yard in a type of carpet material varies with
mean 1.5 flaws per square yard and standard deviation 1.3 flaws per square yard. This population
distribution can not be normal, because a count takes only whole-number values (i.e. this is a discrete
population). An inspector studies 200 square yards of the material, records the number of flaws found per
square yard inspected. Use the central limit theorem to find the approximate probability that the mean
number of flaws exceeds 2 per square yard.




2 - 1.5
P(X > 2) = P  Z >

1.3 



200 

= P(Z > 5.44)
= 0.0000000268
The result indicates that the probability of seeing more than 2 flaws per square yard on average from a
sample of 200 is extremely rare.
In this situation we have a population that is not normally distributed. Regardless of the distribution type
we can still calculate the mean and standard deviation but the 68-95-99.7 rule does not apply anymore.
We are told that µ = 1.5 flaws per sq yd, and that σ = 1.3 flaws per sq yd. Again, you can see that this is
not normally distributed since the smallest value for a measurement is 0 flaws per sq yd, and if we go two
standard deviations to the left 1.5 – 2(1.3) we have negative
flaws per sq yd which is nonsense. Also normal distributions
are continuous and this distribution is discrete, we can only have
whole numbers as our outcomes; 0 flaws/sq yd, 1 flaw/sq yd, 2
flaws/sq yd, and so on. We can not have 1.75flaws/sq yd in one
measurement; when we average several values then we can have
fractional flaws/sq yd.
The situation is we will be looking at 200 sq yds of material, and
we want to know the likely hood that if we looked at blocks of
1yd by 1yd and recorded the flaws in each square yard, and then
averaged all 200 numbers, what is the probability that the average
recorded is greater than 2 flaws/sq yd?
Simulated distribution based on given
information.
Key words - Central Limit Theorem – the theorem is mentioned to bring back to memory the fact that you
will be dealing with the sampling distribution and not the actual population itself. Also the distribution is
approximately normal, according to the theorem, so our calculated probability is also an approximation.
You can see from the histogram, and the normal quantile plot
that the Central Limit Theorem is correct in the fact that the
distribution is very, very close to a normal distribution. The
straight line in the normal quantile plot indicates that it is
extremely close to a normal distribution, so much so that we
can depend on the calculations we are about to make to be
very good approximations.
I want to calculate the probability that my average of 200
numbers exceeds 2 flaws per sq yd. I can see from my
histogram that this is not very likely, since in my 750
simulations not once did this occur.
I used the simulated distribution above to
sample from.
Normal Quantile Plot for the Sampling
Dsitribution Simulation; 750 sample means
averaging 200 values at a time.
1.9
1.8
1.7
1.6
1.5
1.4
1.3
AVERAGES OF 200
VALUES
I conducted a simulation in which I sampled 200 square yards
of material and recorded the flaws in the 200 pieces of 1 yd by
1 yd. I then averaged the 200 values I recorded to get my one
value of the sample mean, x . Now I repeated this procedure
765 more times to obtain the sampling distribution of the
mean, for the 765 values. By doing this many times I am
hoping that the distribution I got by experiment is close to the
theoretical distribution, or at least I get to glimpse what the
theoretical distribution probably looks like.
-4
-2
0
EXPECTED Z-SCORE
2
4