Suppose that the probability that a person books an airline ticket using an online travel website is 0.72. This is from a sample of ten randomly selected people who recently booked an airline ticket. Out of ten randomly selected people, how many would you expect to use an online travel website to book their hotel, give or take how many

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Answer:

The expected value is given by this formula:

[tex]E(X) = np=10*0.72=7.2[/tex]

And the standard deviation for the random variable is given by:

[tex]sd(X)=\sqrt{np(1-p)}=\sqrt{10*0.72*(1-0.72)}=1.420[/tex]

So then we expect about 7.2 people out of 10 that use the online web travel websote to book their hotel.

Step-by-step 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".  

Solution to the problem

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

[tex]X \sim Binom(n=10, p=0.72)[/tex]  

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

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

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

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

The expected value is given by this formula:

[tex]E(X) = np=10*0.72=7.2[/tex]

And the standard deviation for the random variable is given by:

[tex]sd(X)=\sqrt{np(1-p)}=\sqrt{10*0.72*(1-0.72)}=1.420[/tex]

So then we expect about 7.2 people out of 10 that use the online web travel websote to book their hotel.