Question

A survey of consumer finance found that 25.4% of credit-card-holding families hardly ever pay off the balance. Suppose a random sample of 20 credit-card-holding families is taken. Find the probabilities of each of the following results. Source: Statistical Abstract of the United States. 25. Exactly 6 families hardly ever pay off the balance. 26. Exactly 9 families hardly ever pay off the balance. 27. At least 4 families hardly ever pay off the balance. 28. At most 5 families hardly ever pay off the balance.

Answer 1

25. P(6) = 0.1721

26. P(9) = 0.0294

27.
`P(x\geq 4)=0.7868`

28.
`P(x\leq 5)=0.6009`

Explanation:

When we have n identical and independent events with a probability p of success and (1-p) of fail, we have a Binomial distribution. So, the probability that x events are success is calculated as:

`P(x)=(n!)/(x!(n-x)!)*p^(x)*(1-p)^(n-x)`

So, the probability that x families hardly ever pay off the balance follows a binomial distribution and it is calculated as:

`P(x)=(20!)/(x!(20-x)!)*0.254^(x)*(1-0.254)^(20-x)`

where n is 20 and p is 0.254.

Then, the probability that exactly 6 families hardly ever pay off the balance is:

`P(6)=(20!)/(6!(20-6)!)*0.254^(6)*(1-0.254)^(20-6)=0.1721`

The probability that exactly 9 families hardly ever pay off the balance is:

`P(9)=(20!)/(9!(20-9)!)*0.254^(9)*(1-0.254)^(20-9)=0.0294`

The probability that at least 4 families hardly ever pay off the balance is:

`P(x\geq 4)=P(4)+P(5)+...+P(19)+P(20)`

So, using the same equation to find every probability, we get:

`P(x\geq 4)=0.7868`

Finally, the probability that at most 5 families hardly ever pay off the balance is:

`P(x\leq 5)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)\\P(x\leq 5)=0.6009`