Question

An airline has found about 8% of its passengers request vegetarian meals. On a flight with 162 passengers the airline has 12 vegetarian dinners available. What's the probability that it will be short of vegetarian dinners? (hint: Let X be the number of vegetarians. Identify a probability model for X. Write a probability expression for X that reflects "Short of vegetarian dinners" and compute.)

Answer 1

Answer:

The expression is $P(X \geq 13)$.

49.60% probability that it will be short of vegetarian dinners.

Step-by-step explanation:

For each passenger, there are only two possible outcomes. Either they request vegetarian meals, or they do not. The probability of a passenger requesting vegetarian meals is independent from other passengers. So we use the binomial probability distribution to solve this problem.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

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:

$n = 162, p = 0.08$

So

$\mu = E(X) = np = 162*0.08 = 12.96$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)}[ = \sqrt{162*0.08*0.92} = 3.46$

On a flight with 162 passengers the airline has 12 vegetarian dinners available. What's the probability that it will be short of vegetarian dinners?

If there are 13 or more passengers requesting vegetarian meals, the flight will be short. So the expression is $P(X \geq 13)$.

This probability is 1 subtracted by the pvalue of Z when X = 13. So

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

$Z = \frac{13 - 12.96}{3.46}$

$Z = 0.01$

$Z = 0.01$ has a pvalue of 0.5040

1 - 0.5040 = 0.4960

49.60% probability that it will be short of vegetarian dinners.