Please answer C & D only! (if you answered this last time please do not again it

Question

Please answer C & D only! (if you answered this last time please do not again it was incomplete) someone new try please! be clear and show all steps, thank you:

3. In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A shipment is acceptable if at least 440 of the 500 bearings meet the specification Assume that each shipment contains a random sample of bearings a) is the probability that a given shipment is acceptable? b) What is the probability that more than 285 out of 300 shipments are acceptable? c) What is the probability that between 270 and 280 out of 300 shipments are acceptable? d) What is the probability that 280 out of 300 shipments are acceptable?

Explanation / Answer

If 90% of the the bearing meet a thickness specification.
Probability that at least 440 out of 500 meet the specification is 0.9190 (value is obtained using binomial formula = 1 - BINOM.DIST(440,500,0.9,TRUE))

Now p = 0.919, n = 300, q = 1-p
mean = np = 275.7039
std. dev. = sqrt(npq) = 4.7253

(C)
Now p = 0.919, n = 300, q = 1-p
mean = np = 275.7039
std. dev. = sqrt(npq) = 4.7253
P(270<X<280) = P((270-275.7)/4.7253 < z < (280-275.7)/4.7253) = P(-1.2063 < z < 0.91) = 0.70467

(D)
Here we can find the probability by using binomial
P(X=280) = 300C280 * p^280 * q^20 = 0.0594

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