Answer:
0.5819 = 58.19% probability that five or more families have life insurance
Step-by-step explanation:
For each family, there are only two possible outcomes. Either they have insurance, or they do not. The probability of a family having insurance is independent of other families. So we use the binomial probability distribution 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
77% of husband-wife families with kids under 18 years old have life insurance.
This means that [tex]p = 0.77[/tex]
A random sample of six husband-wife families was selected.
This means that [tex]n = 6[/tex]
What is the probability that five or more families have life insurance?
This is
[tex]P(X \geq 5) = P(X = 5) + P(X = 6)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{6,5}.(0.77)^{5}.(0.23)^{1} = 0.3735[/tex]
[tex]P(X = 6) = C_{6,6}.(0.77)^{6}.(0.23)^{0} = 0.2084[/tex]
[tex]P(X \geq 5) = P(X = 5) + P(X = 6) = 0.3735 + 0.2084 = 0.5819[/tex]
0.5819 = 58.19% probability that five or more families have life insurance
