Question

Critical Thinking: Empirical/Quantitative Skills

Answer 1

Answer:

1. 0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. 0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. 0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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

Assume that there is a 0.905 probability that a passenger with a ticket will show up for the  flight.

This means that $p = 0.905$

Also assume that the airline sells 200 tickets

This means that $n = 200$

Question 1:

Exactly, so we can use the P(X = x) formula, to find P(X = 180).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 180) = C_{200,180}.(0.905)^{180}.(0.095)^{20} = 0.0910$

0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. When 200 tickets are sold, calculate the probability that at most 180 passengers show up for the flight.

Now we have to use the approximation.

Mean and standard deviation:

$\mu = E(X) = np = 200*0.905 = 181$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.905*0.095} = 4.15$

Using continuity correction, this is $P(X \leq 180 + 0.5) = P(X \leq 180.5)$, which is the p-value of Z when X = 180.5. Thus

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

$Z = \frac{180.5 - 181}{4.15}$

$Z = -0.12$

$Z = -0.12$ has a p-value of 0.4522.

0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. When 200 tickets are sold, calculate the probability that more than 180 passengers show up for the flight.

Complementary event with at most 180 passengers showing up, which means that the sum of these probabilities is 1. So

$p + 0.4522 = 1$

$p = 1 - 0.4522 = 0.5478$

0.5478 = 54.78% probability that more than 180 passengers show up for the flight.