Question

According to the MIT Airline Data Project, American Airlines controlled 15.5% of the domestic market during a recent year. A random sample of 125 domestic passengers that year was selected. Using the normal

asked Apr 8, 2021 99.8k views

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Birju Vachhani · Answer 1 · 2021-04-13T06:15:33+0000

Answer:

P(20 <X<25)

And we can use the z score given by:

$z = (x-\mu)/(\sigma)$

And if we find the z score for 20 and 25 we got:

z = (20-19.375)/(4.046)=0.154

z = (25-19.375)/(4.046)=1.390

And we have the probability given by:

P(20 <X<25)=P(0.154< Z<1.390)

And we can find this probbaility with this difference:

P(0.154< Z<1.390)=P(Z<1.390) -P(Z<0.154) = 0.9177-0.5612 = 0.3565

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=125, p=0.155)$

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^(n-x)

Where (nCx) means combinatory and it's given by this formula:

nCx=(n!)/((n-x)! x!)

We need to check the conditions in order to use the normal approximation.

$np=125*0.155=19.375 \geq 10$

$n(1-p)=125*(1-0.155)=105.625 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=125*0.155=19.375

$\sigma=√(np(1-p))=√(125*0.155(1-0.155))=4.046$

So then we can approximate the random variable like this:

$X\sim N(\mu = 19.375 , \sigma=4.046)$

And we are interested on this probability: