Question

Assume that when adults with smartphones are randomly​ selected, 53​% use them in meetings or classes. If 7 adult smartphone users are randomly​ selected, find the probability that exactly 5 of them use their smartphones in meetings or classes.

Answer 1

Answer:

$P(X=5)=(7C5)(0.53)^5 (1-0.53)^{7-5}=0.194$

Then the probability that exactly 5 of them use their smartphones in meetings or classes is 0.194

Step-by-step explanation:

Let X the random variable of interest "number of adults with smartphones", on this case we now that:

$X \sim Binom(n=7, p=0.53)$

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:

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

And we want to find this probability:

$P(X=5)$

Using the probability mass function we got:

$P(X=5)=(7C5)(0.53)^5 (1-0.53)^{7-5}=0.194$

Then the probability that exactly 5 of them use their smartphones in meetings or classes is 0.194