Question

A governments department of transportation reported that in 2009, airline A led all domestic airlines in on-time arrivals for domestic flights, with a rate of 0.832. Complete parts (a) through (d)

Using the binomial distribution, what is the probability that in the next six flights, four flights will be on time?

Using the binomial distribution, what is the probability that in the next six flights, all six flights will be on time?

Using the binomial distribution, what is the probability that in the next six flights, at least four flights will be on time?

What are the mean and standard deviation of the number of on-time arrivals? Let n=6

The standard deviation of the number of on-time arrivals is?

Answer 1

Solution:

Given in the question

P(On time arrivals) = 0.832

P(No flight on time arrivals) = 1-0.832 = 0.168

Solution(a)

No. Of flight = 6

We need to calculate P(X=4)

Here we wil use binomial distribution theorem which can be calculated as

P(X=n|N,p) = NCn*(p^n)*((1-p)^(N-n))

P(X=4) = 6C4*(0.832)^4*(0.168)^2 = 0.2029

So there is 20.29% probability that four flights will be on time.

Solution(b)

We need to calculate all 6 flights will be on time

P(X=6) = 6C6*(0.832)^6*(0.168)^0 = 0.3317

So there is 33.17% probability that all 6 flights will on time.

Solution(c)

We need to calculate that atleast 4 flights will be on time

P(X>=4) = P(X=4) + P(X=5) + P(X=6) = 6C4*(0.832)^4*(0.168)^2 + 6C5*(0.832)^5*(0.168)^1 + 6C6*(0.832)^6*(0.168)^0 = 0.2029+0.4019+0.3317 = 0.9364

So there is 93.64% probability that atleast 4 flights are on time.

Solution(d)

Mean = n*p = (0.832*6) = 4.99

Standard deviation = sqrt(n*p*q) = sqrt(6*0.832*0.168) = Sqrt(0.838) = 0.92