Answer Key for Discrete Problems
1.
It is estimated that one out of 10 widgets supplied to a manufacturer is defective.
Suppose a random sample of 20 is chosen from a very large batch of these widgets.
What is the probability that the sample contains no more than one defective widget?
Assume that the number of defectives follows a binomial distribution.
We want P(X = 0 or 1) = P(X = 0) + P(X = 1) = 1*0.1
0
*0.9
20
+ 20*0.1
1
*0.9
19
= 0.1216 + 0.2701
= 0.3917
2.
A manufacturer of heating elements for water heaters ships boxes that contain 100
elements.
A qualitycontrol inspector randomly selects a box in each shipment and
accepts the shipment if there are 5 or fewer defective heating elements in the box.
Assuming that the manufacturer has had a rate of 6% defective items, what is the
probability that the shipment of heating elements will pass the inspection?
Let D = # defectives in a box.
We can model D with a binomial distribution with n =
100, p = 0.06.
We want P(D <= 5)
Use BINOMDIST(5,100,0.06,1) = 0.4407.
Alternatively using P(X = k) = nCkp
k
(1p)
nk
,
P(X = 5) = 100C5*.06
5
*.94
95
= (100*99*98*97*96/5*4*3*2*1)*0.00000078* 0.0027999 = .
16392
P(X = 4) = 100C4*.06
4
*.94
96
= (100*99*98*97/(4*3*2*1)*0.00001296*0.002631928 = .13375
P(X = 3) = 100C3*.06
3
*.94
97
= (100*99*98/3*2*1)*0.000216*0.002474012 = .08641
P(X = 2) = 100C2*.06
2
*.94
98
= (100*99/2*1)*0.0036* 0.00232557 = .04144
P(X = 1) = 100C1*.06
1
*.94
99
= (100/1)*0.06*0.002186 = .01312
P(X = 0) = 100C0*.06
0
*.94
100
=1*0.002* 1 = .00205
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P(X <=5) = sum of above = .16392 + .13375 + .08641 + .04144 + .01312 + .00205 = 0.44069 ~
0.4407
3.
In Canada, Quebec is a key part of the chemical industry.
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 Spring '08
 Clark,B
 Statics, Normal Distribution, Poisson Distribution, Probability theory, Binomial distribution, blemishes, Space Shuttle Challenger

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