A nationwide survey of seniors by the University of Michigan reveals that almost 18.0% disapprove of daily pot smoking. If 8 seniors are selected at random, what is the probability that at least 2 disapprove of daily pot smoking.

Respuesta :

Answer:

[tex] P(X\geq 2)=1- P(X<2)= 1-[P(X=0) +P(X=1)][/tex]

And using the probability mass function we can find the individual probabilities:

[tex]P(X=0)=(8C0)(0.18)^0 (1-0.18)^{8-0}=0.2044[/tex]

[tex]P(X=1)=(8C1)(0.18)^1 (1-0.18)^{0-1}=0.3590[/tex]

And replacing we got:

[tex] P(X\geq 2)=1 -[0.2044 +0.3590]= 0.4366[/tex]

Then the probability that at least 2 disapprove of daily pot smoking is 0.4366

Step-by-step explanation:

Let X the random variable of interest "number of seniors who disapprove of daily smoking ", on this case we now that:

[tex]X \sim Binom(n=8, p=0.18)[/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]

And we want to find this probability:

[tex] P(X\geq 2)=1- P(X<2)= 1-[P(X=0) +P(X=1)][/tex]

And using the probability mass function we can find the individual probabilities:

[tex]P(X=0)=(8C0)(0.18)^0 (1-0.18)^{8-0}=0.2044[/tex]

[tex]P(X=1)=(8C1)(0.18)^1 (1-0.18)^{0-1}=0.3590[/tex]

And replacing we got:

[tex] P(X\geq 2)=1 -[0.2044 +0.3590]= 0.4366[/tex]

Then the probability that at least 2 disapprove of daily pot smoking is 0.4366