Mathematics, 16.03.2020 18:56 wyattgrubb00

An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline sells 105 tickets for a flight that has only 100 seats, what is 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.)

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Answer from: hollimonpat

0.5438 = 54.38% probability that a seat will be available for every person holding a reservation and planning to fly

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

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 = 105, p = 1 - 0.05 = 0.95

I use p = 0.95 because i consider a success a person showing up to the flight. 5% probability that a person misses the flight, so 100-5 = 95% probability that a person shows up to the flight.

For the approximation:

$\mu = E(X) = np = 105*0.95 = 99.75$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.95*0.05} = 2.23$

What is the probability that a seat will be available for every person holding a reservation and planning to fly?

Probability of 100 or less people showing up, which is the pvalue of Z when X = 100. So

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

$Z = \frac{100 - 99.75}{2.23}$

Z = 0.11

Z = 0.11 has a pvalue of 0.5438

0.5438 = 54.38% probability that a seat will be available for every person holding a reservation and planning to fly

Answer from: Quest

Set square i think so

Answer from: Quest

answer: plz put a pic in so i can answer it

step-by-step explanation:

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