# Thread: binomial distribution + approximation

1. ## binomial distribution + approximation

An airline sells 108 tickets for her daily flight from Tokyo to Los Angeles. On average, 4% of customers who have purchased tickets do not turn up. Let X represent the number of customers who do not turn up for this flight.

ii) Find the probability that the mean number of empty seats for each daily flight is at most 4, in a randomly chosen 60-day period.
iii) By using a suitable approximation, find the least number of consecutive days in which the probability of at least 6 customers do not turn up exceeds 0.999.

the answer for ii) is 0.112 and the answer for iii) is 4. But I've spent a very long time working through the question and none of my methods seem correct. If you can explain, please do! Thanks.

2. ## Re: binomial distribution + approximation

X has mean 108*0.04=4.32 and variance 108*0.04*0.96=4.147 So total (T) over 60 days has mean 60*4.32=259.2 and variance 60*4.147=248.82 and standard deviation root248.82=15.77. For mean at most 4 we want T to be at most 240

Using normal approximation P(T<240)=P(Z<(240-259.2)/15.77) That is want Z<-1.218 Using tables this is 1-0.8884=0.112

thanks!