Please help thanks Problem 4 A supplier claims that at most
Please help :( thanks
Problem 4. A supplier claims that at most 5% of the concrete beams fail the strength test. To test the claim you, as a quality-control engineer, randomly select 25 beams and run the strength test. (a) What is the probability that 2 beams in the sample fail the test? (b) What is the probability that at least 4 beams fail the test? (c) What is the probability that between 2 and 5 beams fail the test? (d) What is the probability that at most 2 beams fail the test?Solution
a)
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 = 25
p = the probability of a success = 0.05
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 0.230517651 [ANSWER]
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b)
Note that P(at least x) = 1 - P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 25
p = the probability of a success = 0.05
x = our critical value of successes = 4
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 3 ) = 0.965909399
Thus, the probability of at least 4 successes is
P(at least 4 ) = 0.034090601 [ANSWER]
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c)
Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)
Here,
x1 = 2
x2 = 5
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 25
p = the probability of a success = 0.05
Then
P(at most 1 ) = 0.642375854
P(at most 5 ) = 0.998787039
Thus,
P(between x1 and x2) = 0.356411185 [ANSWER]
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d)
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 25
p = the probability of a success = 0.05
x = the maximum number of successes = 2
Then the cumulative probability is
P(at most 2 ) = 0.872893504 [ANSWER]
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