Answer

Using normal approximation we get:

X ~ N(8641*0.61, 8641*0.61(1-0.61)) = N(5271.01, 2055.6939)

(A)

Less than 5200 defectives means 5199 or less on the discrete scale, but 5199 extends up as far as 5199.5 on the continuous scale. So the approximation is the area under the Normal curve below 5199.5. (This is called the continuity correction)

P(X < 5199.5) = P(Z < (5199.5-5271.01)/√2055.6939) = P(Z < -1.577) = 0.0574


(B)

More than 5400 defectives means 5401 or more on the discrete scale, but 5401 extends down as far as 5400.5 on the continuous scale. So the approximation is the area above the Normal curve above 5400.5.

P(X > 5400.5) = P(Z > (5400.5-5271.01)/√2055.6939) = (P > 2.856) = 0.0021

(C)

P(5199.5 < X < 5400.5) = P(-1.577 < Z < 2.856) = 0.94

(D)

n = 8641, p = 0.61, q = 1-0.61, yes both np and nq are larger than 5. This is needed for a good normal approximation.

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Sk1llz