According to a study, the probability that a randomly selected teenager shopped at a mall at least once during a week was 0.67. Let X be the number of students in a randomly selected group of 50 that will shop at a mall during the next week. (a) Compute the expected value and standard deviation of X. (Round your answer to two decimal places.) Hint [See Example 5.] expected value standard deviation (b) Fill in the missing quantity. (Round your answer to the nearest whole number.) There is an approximately 2.5% chance that or more teenagers in the group will shop at the mall during the next week.

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".

2) Part a

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

[tex]X \sim Binom(n=50, p=0.67)[/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=50*0.67=33.50[/tex]

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

[tex]sd(X)=\sqrt{np(1-p)}=\sqrt{50*0.67*(1-0.67)}=3.32[/tex]

3) Part b

We need to check if we can use the normal approximation , the conditions are:

[tex]np=50*0.67=33.50>10[/tex] and [tex]n(1-p)=50*(1-0.67)=16.5>10[/tex]

So then we can apply the normal approximation to the binomial distribution in our case:

[tex]X \sim N(\mu=33.50,\sigma=3.32)[/tex]

We need on the right tail of the distribution a value a that gives to us 2.5% of the area below and 97.5% of the area above. Both conditions are equivalent.

Let's use the condition [tex]P(X<a)=0.025[/tex], the best way to solve this problem is using the z score with the following formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

So we need a value from the normal standard distribution that accumulates 0.025 of the area on the left and 0.975 on the right. This value on this case is -1.96 and we can founded with the following code in excel:

"=NORM.INV(0.025,0,1)"

If we apply the z score formula to our case we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-33.5}{3.32})=P(Z<-1.96)=0.025[/tex]

So then based on the equalities we have this:

[tex]\frac{a-33.5}{3.32}=-1.96[/tex]

And if we solve for a we got:

[tex]a=(-3.32*1.96) +33.50=26.99[/tex]

There is an approximately 2.5% chance that 27 or more teenagers in the group will shop at the mall during the next week.