Question

Consider the probability that more than 92 out of 146 flights will be on-time. Assume the probability that a given flight will be on-time is 63%.

Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer #1

Solution :

Given that,

p = 0.63

q = 1 - p =1-0.63=0.37

n = 146

Using binomial distribution,

= n * p = 146*0.63=91.98

= n * p * q = 146*0.63*0.37=5.8337

Using continuity correction

,P(x >92 ) = 1 - P(x <92.5 )

= 1 - P((x - ) / < (92.5-91.98) / 5.8337)

= 1 - P(z <0.09 )

Using z table

= 1-0.5359

probability= 0.4641

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