Question

8. Suppose 22% of the eggs sold at a local grocery store that are graded “large” are smaller than that and should be graded “medium.” A random sample of 15 eggs graded large is obtained. Answer the following using the binomial distribution:(Round to 4 (FOUR) decimal places.)

What is the probability that 8 or more of the “large” eggs sampled are really medium-sized?

What is the probability fewer than 3 of the “large” eggs sampled are really medium-sized?

What is the probability that none of the “large” eggs sampled are really medium-sized?

What is the probability that exactly 4 of the “large” eggs sampled are really medium-sized?

What is the probability that all of the “large” eggs sampled are really medium-sized?

What is the probability that 6 or 7 of the “large” eggs sampled are really medium-sized?

Answer 1

n = 15

p = 0.22

P(X = x) = nCx * p^x * (1 - p)^{n - x}

a) P(X > 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= 15C8 * (0.22)^8 * (0.78)^7 + 15C9 * (0.22)^9 * (0.78)^6 + 15C10 * (0.22)^10 * (0.78)^5 + 15C11 * (0.22)^11 * (0.78)^4 + 15C12 * (0.22)^12 * (0.78)^3 + 15C13 * (0.22)^13 * (0.78)^2 + 15C14 * (0.22)^14 * (0.78)^1 + 15C15 * (0.22)^15 * (0.78)^0 = 0.0078

b) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

                  = 15C0 * (0.22)^0 * (0.78)^15 + 15C1 * (0.22)^1 * (0.78)^14 + 15C2 * (0.22)^2 * (0.78)^13 = 0.3269

c) P(X = 0) = 15C0 * (0.22)^0 * (0.78)^15 = 0.0241

d) P(X = 4) = 15C4 * (0.22)^4 * (0.78)^11 = 0.2079

e) P(X = 15) = 15C15 * (0.22)^15 * (0.78)^0 = 0.00001

f) P(X = 6) + P(X = 7) = 15C6 * (0.22)^6 * (0.78)^9 + 15C7 * (0.22)^7 * (0.78)^8 = 0.0826