Question

A motel has a policy of booking as many as 150 guests in a building that holds 140. Past studies indicate that only 85% of booked guests show up for their room. Find the probability that if the motel books 150 guests, not enough seats will be available.

Answer 1

0.0015 = 0.15% probability that if the motel books 150 guests, not enough seats will be available.

Explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`, if
`np \geq 10` and
`n(1-p) \geq 10`.

150 guests booked:

This means that
`n = 150`

85% of booked guests show up for their room.

This means that
`p = 0.85`

Is the normal approximation suitable:

`np = 150(0.85) = 127.5`

`n(1-p) = 150(0.15) = 22.5`

Both greater than 10, so yes.

Mean and standard deviation:

`\mu = E(X) = np = 150*0.85 = 127.5`

`\sigma = √(V(X)) = √(np(1-p)) = √(150*0.85*0.15) = 4.3732`

Find the probability that if the motel books 150 guests, not enough seats will be available.

More than 140 show up, which, using continuity correction, is
`P(X > 140 + 0.5) = P(X > 140.5)`, which is 1 subtracted by the p-value of Z when X = 140.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (140.5 - 127.5)/(4.3732)`

`Z = 2.97`

`Z = 2.97` has a p-value of 0.9985.

1 - 0.9985 = 0.0015.

0.0015 = 0.15% probability that if the motel books 150 guests, not enough seats will be available.