Question

A certain flight arrives on time 87 percent of the time. Suppose 181 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 155 flights are on time.

​(b) at least 155 flights are on time.

​(c) fewer than 170 flights are on time.

​(d) between 170 and 171​, inclusive are on time.

Answer 1

It is given that the flight arrives on time is 87% . ie p-0.87. It is given that n=181. Also, it is given asked to use normal approximatin to Binomial distribution. Therefore X is approximately N(np, npq), where n=181, p=0.87 and q=1-p=0.13.

Ie X~N(157.47,20.4711) Since variance=20.4711,

Since X~N(157.47,20.4711) we know that ~N(0,1).

We need to find the probability that

a.exactly 155 flights are on time. ie P(X=155) =0 since we have approximated teh distribution to a Normal which is a continous distribution and for a continous distribution, probaility at any particular point is zero.

b.at least 155 flights are on time. This is X>=155. That is

  

  

We need to find the probability of . Or in otherwords,

  

ie the probability that at least 155 flights are on time = 0.7075 or 70.75%

​(c) fewer than 170 flights are on time.

That is P(X<170)

  

  

  

The probability of fewer than 170 flights are on time= 0.9972 ie 99.72%

​(d) between 170 and 171​, inclusive are on time.

ie P(170<=X<=171)

ie the probability that between 170 and 171​, inclusive are on time =0.0014 or 0.14%.