1.5k views
3 votes
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.)

1 Answer

5 votes

Answer:

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

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:


√(V(X)) = √(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 = (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 = √(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 = √(V(X)) = √(np(1-p)) = √(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 = (X - \mu)/(\sigma)


Z = (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.

User Russ Hyde
by
7.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories