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Homework #8. Solution.
IE 230
Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers, John
Wiley & Sons, New York, 1999. Chapter 5, Sections 5.6-5.8. Pages 10-11 of the concise notes.
1. A random variable X is standardized by subtracting its mean, µX , and dividing the difference by its
standard deviation, σX . The standardized random variable,
Y=
X − µX
,
σX
h
hhhhhh
has zero mean and unit variance. That is, E(Y) = 0 and V(Y) = 1.
(a) To the nearest integer, the scores on Exam #1 had a mean of 76 and a standard deviation of
11. Compute your standardized Exam #1 score.
-----------------------------------------------------------------------------------------Suppose that your score is x = 80.
Then your standardized score is z = (x − µX ) / σX = (80 − 76) / 11 = 4 / 11.
That is, your score is four-elevenths of a standard deviation above the mean.
-----------------------------------------------------------------------------------------(b) Result: The mean of a standardized random variable is zero. Show that this result is true by
simplifying
X − µX M ∞ I x − µX M
J f (x) dx .
J =
E(Y) = E J
∫ J hhhhhh
σX O X
σX O −∞
L
L
-----------------------------------------------------------------------------------------I
h
hhhhhh
∞
∞
∫ x fX (x) dx − ∫ µX fX (x) dx
−∞
−∞
=
h
hhhhhhhhhhhhhhhhhhhhhhh
=
hhhhhhhhh
σX
E(X) − µX
σX
=0 .
This proof assumed that X is a continuous random variable.
For a discrete random variable, substitute summations for integrals.
For either, be sure that you understand why each step is true.
-----------------------------------------------------------------------------------------(c) Result: The variance of a standardized random variable is one. Show that this result is true by
simplifying
X − µX
V(Y) = V J
σX
L
I
h
hhhhhh
R
X − µX
J = E
J
J
σX
O
L
M
J
I
h
hhhhhh
Q
M
J
O
2
∞
H
J
J
P
=
∫
−∞
x − µX
J
σX
L
I
hhhhhh
2
M
J
fX (x) dx .
O
-----------------------------------------------------------------------------------------∞
=
∫
−∞
x − µX
2
I
M
L
O
fX (x) dx
h
hhhhhhhhhhhhhhhhhh
σ 2X
= V(X) / σ 2X
=1 .
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Schmeiser
Homework #8. Solution.
IE 230
(d) Is your Exam #1 score an example of x or X?
-----------------------------------------------------------------------------------------An observed exam score is a constant, x.
The random variable is X, here the score of a randomly selected student.
-----------------------------------------------------------------------------------------2. (From Problem 5-52.) Let X denote the diameter of the dot produced by a particular printer.
Assume that X is normally distributed with mean µ = 0.002 inch and standard deviation
σ = 0.0004 inch.
(a) Is it possible that X has exactly a normal distribution? Argue why or why not.
-----------------------------------------------------------------------------------------No. A diameter cannot be negative, but the normal range is (−∞, ∞).
-----------------------------------------------------------------------------------------(b) Carefully sketch this normal density function (pdf). Label and scale both axes.
-----------------------------------------------------------------------------------------The horizontal axis has label x; the vertical label f (x) or fX (x).
Draw the bell curve.
Scale the horizontal axis with 0.002 at the center of the bell curve.
The points of inflection are at x = 0.002−+ 0.0004.
The bell curve visually ends at about x = 0.002−+ 3 × 0.0004.
The vertical axis is scaled with zero (at the bottom of the bell curve) and
with 1 / (σ √2ddd
π ), which is approximately 0.4 / σ
(at the top of the bell curve).
-----------------------------------------------------------------------------------------(c) Draw a second horizontal axis under your pdf sketch. Label this axis z and scale it according
to the standardized values z = (x − µ) / σ. (Notice that this scaling will never change,
regardless of the values of µ and σ.)
-----------------------------------------------------------------------------------------Label this horizontal axis z. Scale it with the center of the bell curve
at z = 0, the points of inflection at z =−+ 1, and the curve visually
disappearing at about z =−+ 3.
-----------------------------------------------------------------------------------------(d) Let B denote the event that the dot is "big", defined as having a diameter greater than 0.0026
inch. Write B in terms of X.
-----------------------------------------------------------------------------------------B = "X > 0.0026 inch"
-----------------------------------------------------------------------------------------(e) Shade the area (from Part (b)) corresponding to B. Visually estimate the probability.
-----------------------------------------------------------------------------------------Shade the area under f (x) for all x > 0.0026.
The shaded area should be about 0.05 to 0.10.
-----------------------------------------------------------------------------------------(f) Compute the z-value obtained by standardizing X at X = 0.0026 inch. Show this value in the
sketch of Part (b).
-----------------------------------------------------------------------------------------z = (x − µ) / σ = (0.0026 − 0.0020) / 0.0004 = 1.5
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Schmeiser
Homework #8. Solution.
IE 230
(g) Compute P(B) using Table II of Appendix A. Does it match your shaded area from Part (e)?
If not, correct your sketch or your calculation.
-----------------------------------------------------------------------------------------P(B) = P(X > 0.0026 in) = P(Z > 1.5)
= 1 − P(Z ≤ 1.5) = 1 − 0.933193 = 0.066807,
where the value 0.033193 is from the cumulative standard normal table.
-----------------------------------------------------------------------------------------(h) Use MSExcel to compute the answer to Part (g). Handwrite the answer and your MSExcel
cell code.
-----------------------------------------------------------------------------------------The MSExcel cell code "1 − NORMDIST(0.0026, 0.002, 0.0004, 1)"
yields the answer 0.066807229 .
-----------------------------------------------------------------------------------------(i) Compute P(X ≤0.0026) inch using your answer to Part (g).
-----------------------------------------------------------------------------------------P(X ≤ 0.0026) = 1 − P(X > 0.0026) = 1 − 0.066807 = 0.933193.
-----------------------------------------------------------------------------------------(j) Let p denote your answer to Part (i). Use zp ≈ (p .135 − (1−p).135 ) / .1975 to approximate the
pth quantile of Z. Show this value in your sketch of Part (b). If it is not close to the z value
from part (f), check your work.
-----------------------------------------------------------------------------------------The simple standard normal inverse cdf approximation
zp ≈ (p .135 − (1−p).135 ) / .1975 yields zp ≈ 1.502393 for p = 0.933193.
-----------------------------------------------------------------------------------------3. (From Problem 5-53). Let X denote the weight of a randomly selected running shoe. Assume that
X is normally distributed with a mean of 12 ounces and a standard deviation of 0.5 ounces.
(a) Carefully sketch this normal density function (pdf). Label and scale both axes. Label and
scale the z axis.
-----------------------------------------------------------------------------------------The horizontal axis has label x; the vertical label f (x) or fX (x).
Draw the bell curve.
Scale the horizontal axis with 12 ounces at the center of the bell curve.
The points of inflection are at x = 12−+ 0.5 ounces.
The bell curve visually ends at about x = 12−+ 3 × 0.5 ounces.
The vertical axis is scaled with zero (at the bottom of the bell curve)
and with approximately 0.4 / σ =0.8
(at the top of the bell curve).
-----------------------------------------------------------------------------------------(b) Carefully sketch the cdf. Label and scale both axes.
-----------------------------------------------------------------------------------------The horizontal axis has label x; the vertical label F (x) or FX (x).
Draw the S curve. (Like the bell curve, this S shape never changes.)
Scale the horizontal axis with 12 ounces at the center of the S curve.
The S curve visually ends at about x = 12−+ 3 × 0.5 ounces.
The vertical axis is scaled with zero (at the bottom of the S curve) and
with one (at the top of the bell curve).
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Schmeiser
Homework #8. Solution.
IE 230
(c) Let L denote the event that the shoe is light, defined as weighing less than 11 ounces. Show
P(L) in the sketches of Parts (b) and (c). Do the two sketches suggest about the same
value? If not, draw the sketches with more care. (If you wish, you can check your answer
by using Table II and/or MSExcel.)
-----------------------------------------------------------------------------------------In Part (b), shade the area under the pdf to the left of 11 ounces.
In Part (c), mark 11 ounces on the horizontal axis, go up to the cdf,
then to the left, hitting the vertical axis at P(L) = P(X < 11).
-----------------------------------------------------------------------------------------(d) What is the weight (in ounces) that 95% of the shoes exceed? Method 1. Solve graphically
using your cdf sketch to find x .05 = F −1 (.05). (We call x .05 the .05 quantile of X.)
-----------------------------------------------------------------------------------------Shade the rightmost 95% of the area under the pdf.
The answer, x 0.05 , is the left edge of the shading.
We know that about 95% of any normal pdf lies within two standard
deviations from the mean. Therefore, about 97.5% lies to the right of
the point two standard deviations below the mean,
which here is 12 − 2 × 0.5 = 11 ounces.
Therefore, the answer x 0.05 is a bit greater than 11 ounces.
-----------------------------------------------------------------------------------------(e) What is the weight (in ounces) that 95% of the shoes exceed? Method 2: Approximate z .05
and then compute x .05 = µ + σ z .05 . (Check your answer with Part (e).)
-----------------------------------------------------------------------------------------Approximately, z 0.05 = −1.65.
Our answer is x 0.05 = µ + σ z 0.05 = 12 + 0.5 × (−1.65) = 11.18 ounces.
-----------------------------------------------------------------------------------------(f) What is the weight (in ounces) that 95% of the shoes exceed? Method 3: Obtain z .05 from
Table II and proceed as in Part (e). (Check your answer with Parts (d) and (e).)
-----------------------------------------------------------------------------------------Again we want z 0.05 .
The standard normal cdf table yields −1.65 < z 0.05 < −1.64.
Interpolating (which is probably overkill) yields z 0.05 = −1.645.
Then the answer is x 0.05 = µ + σ z 0.05 = 12 + 0.5 × (−1.645) = 11.178 ounces.
-----------------------------------------------------------------------------------------(g) What is the weight (in ounces) that 95% of the shoes exceed? Method 4: Use MSExcel. Give
the answer and the cell code.
-----------------------------------------------------------------------------------------From MSExcel’s NORMSINV function, z 0.05 = −1.644853.
(You could also use NORMINV with µ = 0 and σ = 1.)
Then the answer is
x 0.05 = µ + σ z 0.05 = 12 + 0.5 × (−1.644853) = 11.17757 ounces.
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Schmeiser
Homework #8. Solution.
IE 230
4. The normal distribution is often appropriate when the random variable is a sum of many other
random variables. Consider a golfer’s score for eighteen holes. Let Xi denote the number of
strokes required for hole i for i = 1, 2,..., 18. Then the golfer’s score is X = Σ 18 Xi .
i =1
(a) Is X 6 or X better modeled by a normal distribution?
-----------------------------------------------------------------------------------------The Central Limit Theorem, which we will study later, says that the
sum of random variables is approximately normally distributed.
Therefore, X will be closer to a normal distribution than will X 6 .
Another argument is that the golf score for any one hole is very
discrete, ranging from about two to only (one hopes) about seven. The golf
score for an entire round is still integer, but the range is (again, one
hopes) about twenty strokes, so the discreteness is not so important.
-----------------------------------------------------------------------------------------(b) Suppose that a group of four people play a round of golf. Which would be closer to normally
distributed: an individual’s score or the total score?
-----------------------------------------------------------------------------------------The total score, for the same reasons as Part (a).
-----------------------------------------------------------------------------------------(c) Suppose that a particular golfer, on a particular course, has a mean score 84 strokes, with a
standard deviation of 3 strokes. Approximate, using a normal distribution, the probability
that the golfer shoots (exactly) a 78.
-----------------------------------------------------------------------------------------Let X = "number of strokes on a randomly selected day" (for this golfer and course).
Apply the continuity correction to obtain P(X=78) = P(77.5 < X ≤ 78.5).
To use either a standard normal cdf table or MSExcel, we need to convert to an
expression
containing
only
cdf
values:
P(77.5 < X < 78.5) = P(X < 78.5) − P(X < 77.5).
You can use MSExcel:
NORMDIST(78.5,84,3,1) − NORMDIST(77.5,84,3,1) = 0.0334 − 0.0151 = 0.0183.
Alternatively, you can use the standard normal cdf table:
First convert to standard normal values.
P(X ≤ 78.5) −hh
P(X ≤ 77.5) = P(Z
≤ (78.5 − 84) / 3)−P(Z ≤ (77.5 − 84) / 3)
hh
= P(Z ≤ −1.833) − P(Z ≤ −2.166)
With interpolation,
the tablehhyields
hh
P(Z ≤ −1.833) − P(Z ≤ −2.166) = 0.0333 − 0.0151 = 0.0182.
-----------------------------------------------------------------------------------------5. (From Problem 5-61, modified.) A supplier ships a lot of 1000 electrical connectors. A sample of
25 is selected at random, with replacement. Assume that the lot contains 100 defectives. Let X
denote the number of defectives in the sample.
(a) What is the distribution of X, including the parameter values?
-----------------------------------------------------------------------------------------Binomial, with n = 25 and p = 100 / 1000 = 0.1.
(The distribution is binomial because sampling is with replacement.
The distribution would be hypergeometric if sampling were without replacement.)
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Schmeiser
Homework #8. Solution.
IE 230
(b) What is the mean and standard deviation of X?
-----------------------------------------------------------------------------------------µX = np = (25)(0.1) = 2.5.
dddddd
ddddddddd = 1.5.
σX = √np
(1−p) = √(25)(0.1)(0.9)
-----------------------------------------------------------------------------------------(c) Write the exact expression for P(X = 2).
-----------------------------------------------------------------------------------------The binomial pmf.
I
25 M
P(X = 2) = J J0.12 (1 − 0.1)25−2
2 O
L
(25)(24)
(Simplifying yields P(X = 2) = hhhhhhhh 0.12 0.923 = 0.2659.)
2
-----------------------------------------------------------------------------------------(d) Approximate P(X = 2) using the normal distribution.
-----------------------------------------------------------------------------------------P(X = 2) = P(1.5 < X ≤ 2.5), the continuity correction.
Converting to cdf expressions yields P(X ≤ 2.5) − P(X ≤ 1.5).
The normal approximation assumes that X is normal with µ = 2.5 and σ = 1.5.
Trivially P(X ≤ 2.5) = 0.5.
Evaluate P(X ≤ 1.5) graphically, with MSExcel, or with a standard normal cdf table.
Graphically: Approximately, P(X ≤ 1) = 0.16 because 68% of the
normal pdf area is within one standard deviation of the mean.
Therefore, 32% is outside. Half of the area outside one standard
deviation is in the left tail.
So P(X < 1.5) is about 0.2 (or whatever it looks like in your plot).
MSExcel: NORMDIST(1.5,2.5,1.5,1) = 0.2525
Standard normal table:
P(X < 1.5) = P(Z < (1.5−2.5) / 1.5) = P(Z < −2 / 3).
Interpolating from the table yields P(Z < 2 / 3) = 0.2525.
Taking the difference yields the approximation P(X = 2) = 0.5 − 0.2525 = 0.2475.
(Notice that the approximation is pretty good, even though np < 5, violating
a typical guideline for when the normal approximation can be used.)
-----------------------------------------------------------------------------------------(e) Sketch the normal pdf approximation, including your answer to Part (d).
-----------------------------------------------------------------------------------------The horizontal axis has label x; the vertical label f (x) or fX (x).
Draw the bell curve.
Scale the horizontal axis with 2.5 at the center of the bell curve.
The points of inflection are at x = 1 and x = 4.
The bell curve visually ends at about x = 2.5−+ 3 × 1.5,
so it extends substantially below zero.
The vertical axis is scaled with zero (at the bottom of the bell curve)
and with approximately 0.4 / σ =0.4 / 1.5 = 0.27.
(at the top of the bell curve).
Shade the area between x = 1.5 and x = 2.5.
This shaded area is the normal approximation to P(X = 2).
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Schmeiser