Question

According to a 2018 survey by Bankrate, 20% of adults in the United States save nothing for retirement (CNBC website). Suppose that sixteen adults in the United States are selected randomly. What is the probability that three or less of the selected adults have saved nothing for retirement

Answer 1

Answer:

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they save nothing for retirement, or they save something. The probability of an adult saving nothing for retirement is independent of any other adult. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

20% of adults in the United States save nothing for retirement (CNBC website).

This means that $p = 0.2$

Suppose that sixteen adults in the United States are selected randomly.

This means that $n = 16$

What is the probability that three or less of the selected adults have saved nothing for retirement?

This is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{16,0}.(0.2)^{0}.(0.8)^{16} = 0.0281$

$P(X = 1) = C_{16,1}.(0.2)^{1}.(0.8)^{15} = 0.1126$

$P(X = 2) = C_{16,2}.(0.2)^{2}.(0.8)^{14} = 0.2111$

$P(X = 3) = C_{16,3}.(0.2)^{3}.(0.8)^{13} = 0.2463$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0281 + 0.1126 + 0.2111 + 0.2463 = 0.5981$

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement