Suppose you will play 64 independent rounds of a game and have a one in five chance to win each round. Use a normal approximation to estimate the chance you will win at most 10 rounds.
Solution:
We are given
n = 64
p = 1/5 = 0.20
q = 1 – 0.20 = 0.80
Mean = np = 64*0.20 = 12.8
SD = sqrt(npq) = sqrt(64*0.20*0.80) = 3.2
We have to find P(X≤10) = P(X<10.5) (continuity correction)
Z = (X – mean)/SD
Z = (10.5 - 12.8)/3.2
Z = -0.71875
P(Z<-0.71875) = P(X≤10) = 0.236147
(by using z-table)
Required probability = 0.236147
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