Let X = number of flaws on an electroplated automobile grill. Its distribution i
ID: 3155865 • Letter: L
Question
Let X = number of flaws on an electroplated automobile grill. Its distribution is modeled by the following PMF:
1. What is the probability that there are at most 1 flaw on the grill, i.e. P(X less than or equal to 1)?
2. Let Y be total number of grills with at most 1 flaw in a random sample of 100 grills. What is the exact distribution of Y ? Please specify name and parameter values
for distribution of Y .
Find the probability that no grills from the sample of 100 has at most 1 flaw, using the exact distribution.
4. What is the approximate distribution of Y ? Please specify name and parameter values for distribution of Y .
5. Find the probability that less than 85 grills have at most 1 flaw, using the approximate distribution of Y .
x 0 1 2 3 p(x) 0.8 0.1 0.05 0.05Explanation / Answer
a)
P(x<=1) = P(0) + P(1) = 0.8 + 0.1 = 0.9 [ANSWER]
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2.
It is a binomial distribution with n = 100, p = 0.9.
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3.
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 100
p = the probability of a success = 0.9
x = the number of successes = 0
Thus, the probability is
P ( 0 ) = 1*10^-100 [ANSWER]
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4.
It is approximately a normal distribution with
u = mean = np = 90
s = standard deviation = sqrt(np(1-p)) = 3 [ANSWER]
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5.
We first get the z score for the critical value:
x = critical value = 84.5
u = mean = np = 90
s = standard deviation = sqrt(np(1-p)) = 3
Thus, the corresponding z score is
z = (x-u)/s = -1.833333333
Thus, the left tailed area is
P(z < -1.833333333 ) = 0.033376508 [ANSWER]
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