Question

An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be less than 20 no-shows. Use the normal distribution to approximate the binomial distribution.

Answer 1

Answer:

$[tex]P(X$

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=150, p=0.15)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We need to check the conditions in order to use the normal approximation.

$np=150*0.15=22.5 \geq 10$

$n(1-p)=150*(1-0.15)=127.5 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=150*0.15=22.5$

$\sigma=\sqrt{np(1-p)}=\sqrt{150*0.15(1-0.15)}=4.373$

So then our random variable can be distributed on this way:

$X \sim N(\mu =22.5, \sigma=4.373)$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability on this way:

$P(Z$