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Consider the probability that greater than 99 out of 158 flights will be on-time. Assume the probability that a given flight will be on-time is 64%.
Approximate the probability using the normal distribution. Round your answer to four decimal places.

p = 0.64
    µ = 0.64 * 158 = 101.12
             σ = sqrt(0.64*0.36*158) =  6.034
    p(X > 99), using continuity correction factor p(X > 99+0.5)
    z=(99.5-101.12)/6.034= -0.27 
    P(Z < -0.27) = 0.3936
    P(X > 99) =  1 – 0.3936 = 0.6064
 

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