Homework SourseUse the following to answer questions 48 A garment manufactu
Use the following to answer questions 4-8:
A garment manufacturer knows that the number of flaws per square yard in a particular type of cloth that he purchases varies with an average of 1.3 flaws per square yard. The count X of flaws per square yard can be modeled by a Poisson distribution.
.
Using the Poisson probability formula, the probability of getting exactly two flaws in a randomly selected yard of cloth is
a)
0.2169.
b)
0.3012.
c)
0.3614.
d)
0.6626.
.
Using the Poisson probability formula, the probability of getting at most one flaw in a randomly selected yard of cloth is
a)
0.2169.
b)
0.3012.
c)
0.3614.
d)
0.6626.
.
The standard deviation of the number of flaws in a randomly selected yard of cloth is
a)
1.14.
b)
1.30.
c)
1.61.
d)
1.69.
.
The mean number of flaws in five square yards of this cloth is
a)
0.26.
b)
1.14.
c)
1.30.
d)
6.50.
The probability of zero flaws in a square yard is 0.3012. A particular garment requires three square yards of cloth to make. The probability of zero flaws in the three square yards of material is
a)
0.0273.
b)
0.0907.
c)
0.1004.
d)
0.9036.
| . | Using the Poisson probability formula, the probability of getting exactly two flaws in a randomly selected yard of cloth is | |
| a) | 0.2169. | |
| b) | 0.3012. | |
| c) | 0.3614. | |
| d) | 0.6626. |
Solution
1.
Note that the probability of x successes out of n trials is
P(x) = u^x e^(-u) / x!
where
u = the mean number of successes = 1.3
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 0.230289365 [OPTION A is closest, please report this problem to your instructor.]
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2.
Using a cumulative poisson distribution table or technology, matching
u = the mean number of successes = 1.3
x = the maximum number of successes = 1
Then the cumulative probability is
P(at most 1 ) = 0.626823124 [ANSWER, OPTION D is closest, please report this problem.]
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3.
standard deviation = sqrt(mean) = sqrt(1.3) = 1.140175425 [ANSWER, OPTION A]
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4.
As in 1 yd there is 1.3, then in 5 yd, there are 1.3*5 = 6.5. [ANSWER OPTION D]
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5.
P(all three have no flaw) = 0.3012^3 = 0.027325298 [ANSWER, OPTION A]
