Question

According to a Pew Research survey, about 27% of American adults are pessimistic about the future of marriage and the family. This is based on a sample, but assume that this percentage is correct for all American adults. Using a binomial model, what is the probability that, in a sample of 20 American adults, 25% or fewer of the people in the sample are pessimistic about the future of marriage and family?

Answer 1

Answer:

P(X≤5)=0.5357

Step-by-step explanation:

Using the binomial model, the probability that x adults from the sample, are pessimistic about the future is calculated as:

$P(x)=\frac{n!}{x!(n-x)!} *p^{x}*(1-p)^{n-x}$

Where n is the size of the sample and p is the probability that an adult is pessimistic about the future of marriage and family. So, replacing n by 20 and p by 0.27, we get:

$P(x)=\frac{20!}{x!(20-x)!}*0.27^{x}*(1-0.27)^{20-x}$

Now, 25% of 20 people is equal to 5 people, so the probability that, in a sample of 20 American adults, 25% or fewer of the people are pessimistic about the future of marriage and family is equal to calculated the probability that in the sample of 20 adults, 5 people of fewer are pessimistic about the future of marriage and family.

Then, that probability is calculated as:

P(X≤5)= P(1) + P(2) + P(3) + P(4) + P(5)

Where:

$P(0)=\frac{20!}{0!(20-0)!}*0.27^{0}*(1-0.27)^{20-0}=0.0018$

$P(1)=\frac{20!}{1!(20-1)!}*0.27^{1}*(1-0.27)^{20-1}=0.0137$

$P(2)=\frac{20!}{2!(20-2)!}*0.27^{2}*(1-0.27)^{20-2}=0.0480\\P(3)=\frac{20!}{3!(20-3)!}*0.27^{3}*(1-0.27)^{20-3}=0.1065\\P(4)=\frac{20!}{4!(20-4)!}*0.27^{4}*(1-0.27)^{20-4}=0.1675\\P(5)=\frac{20!}{5!(20-5)!}*0.27^{5}*(1-0.27)^{20-5}=0.1982$

Finally, P(X≤5) is equal to:

P(X≤5) = 0.0018+0.0137 + 0.0480 + 0.1065 + 0.1675 + 0.1982

P(X≤5) = 0.5357