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Homework #8. Spring 2001.
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 (through "continuity correction") of the concise notes.
1. (From Problem 5-45, M&R) The fill volume, X , of an automated filling machine used to
fill cans of carbonated beverage is normally distributed with a mean of 12.4 fluid ounces
and a standard deviation of 0.1 fluid ounce.
(a) Is it possible that X has exactly a normal distribution? Argue why or why not.
---------------------------------------------------------------------No. The normal-distribution lower bound is −∞.
---------------------------------------------------------------------(b) Carefully sketch this normal density function (pdf). Label and scale both axes.
---------------------------------------------------------------------Label the horizontal axis with x (or other dummy variable) ounces.
Sketch a bell curve.
Scale the horizontal axis with 12.4 at the bell’s center and 12.1 and 12.7
at the bell’s edges. The points of inflection should be at 12.3 and 12.5.
Label the vertical axis f X (x ) (or substitute your dummy variable).
dd
Scale the axis with zero at the bell’s bottom and 1 / 0.1× √d2π
at the bell’s highest point.
---------------------------------------------------------------------(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 the horizontal axis with z .
Scale it with 0 at the bell’s center and -3 and +3 at the bell’s edges.
The points of inflection should be at -1 and +1.
---------------------------------------------------------------------(d) Let U denote the event that the can is under filled (that is, it contains less than 12
fluid ounces, the amount shown on the label. Write U in terms of X .
---------------------------------------------------------------------U = "X < 12"
---------------------------------------------------------------------(e) Shade the area (from Part (b)) corresponding to P( U ). Visually estimate the
probability.
---------------------------------------------------------------------Shade the area under f X to the left of x = 12 ounces.
---------------------------------------------------------------------(f) Compute the z -value obtained by standardizing X at X = 12 fluid ounces. Show this
value in the sketch of Part (b).
---------------------------------------------------------------------z = (x − µ) / σ = (12 − 12.4) / 0.1 = 4
Mark z = 4 on the second horizontal axis.
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Schmeiser
Homework #8. Spring 2001.
Solution
IE 230
(g) Use MSExcel’s "normdist" to compute the value of P( U ). Handwrite your MSExcel
cell code and the answer.
---------------------------------------------------------------------−5
3.17×10
normdist(12,12.4,.1,1)
---------------------------------------------------------------------(h) Determine the specifications that are symmetric about the mean that include 99% of
all cans. (That is, what interval ( a , b ) has a center at 12.4 and satisfies
P( a < X < b ) = 0.99?)
---------------------------------------------------------------------Let c satisfy a = µ − c and b = µ + c . Then
0.99 = P( a < X < b ) = P( µ − c < X < µ + c ) = P( 12.4 − c < X < 12.4 + c )
implies that 0.995 = P (X < 12.4 + c )
implies that 0.995 = P ((X − µ) / σ < (12.4 + c ) − µ) / σ )
implies that 0.995 = P ((Z < (12.4 + c ) − 12.4) / 0.1 )
implies that 0.995 = P ((Z < 10c )
implies that z .995 = 10c
implies that 2.576 ≈ 10c (from Table II or "normsinv(.995)" of MSExcel)
implies that c ≈ 0.258←
←
----------------------------------------------------------------------
2. (From Problem 5-61, modified.) A supplier ships a lot of 1000 electrical connectors. A
sample of 25 is selected at random, without 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?
---------------------------------------------------------------------X
∼ hypgeodist ( N
= 1000, n = 25, K = 100 )
---------------------------------------------------------------------(b) What is the mean and standard deviation of X ?
---------------------------------------------------------------------Let p = K / N = 100 / 1000 = 0.1. Then
E(X ) = np = 25(100 / 1000) = 2.5 connectors ←
2
σX = np (1−p )(N − n ) / (n − 1) = 25×0.1×(1 − 0.9)×(1000 − 25) / (1000 − 1) ≈ 2.25
Therefore sX = √dddd
2.25 = 1.5 connector ←
---------------------------------------------------------------------(c) Write the exact expression for P( X = 2 ). (If you wish, you can compute the value
manually or using the MSExcel function "hypgeodist".)
---------------------------------------------------------------------Because X is hypgeodist
P( X = 2 ) = C 2 C 23 / C 25 ←
≈ 0.268 from "hypergeodist" in MSExcel.
100
900
1000
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Schmeiser
Homework #8. Spring 2001.
Solution
IE 230
(d) What binomial-distribution parameter values would you use to approximate the
distribution of X ?
---------------------------------------------------------------------n = 25 and p = K / N = 100 / 1000 = 0.1
---------------------------------------------------------------------(e) Compute the binomial approximation for P( X = 2 ).
---------------------------------------------------------------------Here x = 2, n = 25, and p = 0.1. Therefore
n −x
P( X = 2 ) ≈ Cx p (1 − p )
n x
= C 2 0.1 (1 − 0.1)
25
2
25−2
≈ 300×0.01×0.089 = 0.266←
←
---------------------------------------------------------------------(f) Compute the normal approximation for P( X = 2 ).
---------------------------------------------------------------------The mean and standard deviations are µ = 2.5 and σ = 1.5. Therefore
P( X = 2 ) = P( 1.5 < X < 2.5 ) = P( X ≤ 2.5 ) − P( X ≤ 1.5 )
= P( (X − µ) / 1.5 < (2.5−2.5) / 1.5 ) − P( (X − 2.5) / 1.5 < (1.5−2.5) / 1.5 )
= P( Z < 0 ) − P( Z < −2 / 3 )
≈ 0.5 − 0.252 = 0.248←
←
---------------------------------------------------------------------3. The normal distribution is often appropriate when the random variable is a sum of many
other random variables. To rationalize the normal approximation to the binomial
distribution, let’s think of a binomial random variable X as a sum of n indicator random
variables. In particular, let Xi = 1 if the i th Bernoulli trial is a success, and Xi = 0
otherwise, for i = 1, 2,..., n . Then X = Σin=1 Xi .
(a) For n > 1, is X 1 or X better approximated by a normal distribution? (For n = 1?)
---------------------------------------------------------------------X is closer to having a normal distribution.
For n = 1, both X 1 and X have the same distribution because X 1 = X .
---------------------------------------------------------------------(b)
Consider
X
the
event
120 ≤ X ≤ 150
for
the
binomial
random
variable
∼ binomial ( n , p ). Using the continuity correction, write the approximating
event, in terms of Y , a normal random variable. (Notice that your answer does not
depend upon the value of the binomial-distribution parameters n and p or the
normal-distribution parameters µY and σY2 .)
---------------------------------------------------------------------119.5 < X < 150.5
---------------------------------------------------------------------(c) Suppose that X ∼ binomial ( n =200, p = 0.75 ). What value of µY and σY should be
used for the approximating normal distribution? (Notice that your answer does not
depend upon the particular event to be approximated.)
---------------------------------------------------------------------µY = np = 200×0.75 = 150←
←
σY = √ddddddddd
np (1 − p ) = √dddddddddddddddddd
200×0.75×(1 − 0.75) = √dddd
37.5 ≈ 6.12←
←
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Schmeiser
Homework #8. Spring 2001.
Solution
IE 230
(d) Sketch the normal density of Part (c). Shade the probability of the event of Part (b).
Visually estimate the probability.
---------------------------------------------------------------------Label the horizontal axis with x .
Sketch the bell curve over the axis.
Scale the horizontal axis with 150 at the bell’s center.
The edges of the bell should be at about 150−+ 3×6.12
and the points of inflection at about 150−+ 6.12.
Label the vertical axis with f X (x ).
---------------------------------------------------------------------(e) Use the MSExcel function "binomdist" to compute the probability of Part (b).
---------------------------------------------------------------------P( 120 ≤ X ≤ 150 ) = P( X ≤ 150 ) − P( X ≤ 119 )
= binomdist (150,200,.75,1) − binomdist (119,200,.75,1)
≈ 0.527124 − 0.000001
= 0.527123←
←
---------------------------------------------------------------------(f) Use the MSExcel function "normdist" to compute the normal approximation to the
probability of Part (b). (Use the function at the lower bound and at the upper
bound; then take the difference.)
---------------------------------------------------------------------P( 120 ≤ X ≤ 150 ) = P( 119.5 ≤ X ≤ 150.5 )
= P( X ≤ 150.5 ) − P( X ≤ 119.5 )
= normdist (150.5,150,6.12,1) − normdist (119.5,150,6.12,1)
≈ 0.532557 − 0.0000003
≈ 0.536←
←
---------------------------------------------------------------------(g) Check your normal-approximation answer from Part (f) by using the MSExcel
function "norminv".
---------------------------------------------------------------------norminv (.532557,150,6.12) = 150.499996
----------------------------------------------------------------------
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Schmeiser