Deoae whether you can use the normal approximate the binomial distribution, you

Question

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Explanation / Answer

Here p = 0.61, n = 100 and q = 0.39

np = 0.61*100 = 61
nq = 39

As np>= 5 and nq >= 5, we can apply normal approximation.

mean = np = 61
std. dev. = sqrt(npq) = sqrt(100 * 0.61 * 0.39) = 4.8775

(a)
P(X=64) = P(X<64) - P(X<63) = P(z<(64-61)/4.8775) - P(z<(63-61)/4.8775) = P(z<0.6151) - P(z<0.41) = 0.7308 - 0.6591 = 0.0716

The graph which indicates this probabilty is given as option (A)

(b)
P(X>=64) = P(z>(64-61)/4.8775) = P(z>0.6151) = 0.2693

Option (C) gives the appropriate graph for this

(c)
P(X<64) = P(z<(64-61)/4.8775) = P(z<0.6151) = 0.7308

Option (A) is the appropriate graph

(D)
There is no unusual event (option D)

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