Question

An airline finds that 6% of the people who make reservations on a certain flight do not show up for the flight. If the airline sells 190 tickets for a flight with only 185 seats, use the normal approximation to the binomial distribution to find the probability that a seat will be available for every person holding a reservation and planning to fly. (Round your answer to four decimal places.)

Answer 1

Answer:

0.9826 = 98.26% probability that a seat will be available for every person holding a reservation and planning to fly.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses 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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, 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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

6% of the people who make reservations on a certain flight do not show up for the flight. So 100-6 = 94% show up, which means that $p = 0.94$

190 tickets, so $n = 190$

So

$\mu = E(X) = np = 190*0.94 = 178.6$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{190*0.94*0.06} = 3.27$

Find the probability that a seat will be available for every person holding a reservation and planning to fly.

At most 185 people show up.

Using continuity correction, we have to find $P(X \leq 185 + 0.5) = P(X \leq 185.5)$, which is the pvalue of Z when X = 185.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{185.5 - 178.6}{3.27}$

$Z = 2.11$

$Z = 2.11$ has a pvalue of 0.9826

0.9826 = 98.26% probability that a seat will be available for every person holding a reservation and planning to fly.