DO NOT NEED ALL PARTS ANSWERED!! For C i got: .78626 and for E I got .7862591 so
ID: 3129594 • Letter: D
Question
DO NOT NEED ALL PARTS ANSWERED!! For C i got: .78626 and for E I got .7862591 so basically the same. Please help me figure out if these are correct and how to anser F and G THANK YOU SO MUCH IN ADVANCE!!
An unnoticed mechanical failure has caused one-fourth of a machine shop’s production of 10000 pistol firing pins to be defective. A random sample of 25 firing pins was drawn from the population.
a.) Explain why this random variable has a binomial distribution? (2 points)
b.) What are the mean and the standard deviation of the random variable? (2 points)
c.) Find P(X > 4). Use the binomcdf function on your calculator. (2 points)
d.) Is it appropriate to use the normal approximation to the binomial for this problem? Support your answer with numbers and a test. (3 points)
e.) Using the normal approximation with the continuity correction, calculate the probability that you’ll observe more than 4 defective firing pins in the random sample of 25. Indicate the X-values and the z-scores involved. (3 points)
f.) How does your answer compare with the one you have in part c.)? (2 points)
g.) How do you account for the difference in parts c.) and e.)? (2 points )
Explanation / Answer
c)
Note that P(more than x) = 1 - P(at most x).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 25
p = the probability of a success = 0.25
x = our critical value of successes = 4
Then the cumulative probability of P(at most x) from a table/technology is
P(at most 4 ) = 0.213740923
Thus, the probability of at least 5 successes is
P(more than 4 ) = 0.786259077 [ANSWER]
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e)
We first get the z score for the critical value:
x = critical value = 4 + 0.5 = 4.5 [continuity corrected]
u = mean = np = 6.25
s = standard deviation = sqrt(np(1-p)) = 2.165063509
Thus, the corresponding z score is
z = (x-u)/s = -0.808290377
Thus, the left tailed area is
P(z < -0.808290377 ) = 0.79053828 [ANSWER]
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F)
The answer in part c is slightly different to that in part e.
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g)
The difference is basically because part e is just an approximation. The binomial distribution is strictly not continuous, so we cannot expect them to be exactly equal.
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