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.

Respuesta :

Answer:

25. P(6) = 0.1721

26. P(9) = 0.0294

27. [tex]P(x\geq 4)=0.7868[/tex]

28. [tex]P(x\leq 5)=0.6009[/tex]

Step-by-step 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:

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

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

[tex]P(x)=\frac{20!}{x!(20-x)!}*0.254^{x}*(1-0.254)^{20-x}[/tex]

where n is 20 and p is 0.254.

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

[tex]P(6)=\frac{20!}{6!(20-6)!}*0.254^{6}*(1-0.254)^{20-6}=0.1721[/tex]

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

[tex]P(9)=\frac{20!}{9!(20-9)!}*0.254^{9}*(1-0.254)^{20-9}=0.0294[/tex]

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

[tex]P(x\geq 4)=P(4)+P(5)+...+P(19)+P(20)[/tex]

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

[tex]P(x\geq 4)=0.7868[/tex]

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

[tex]P(x\leq 5)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)\\P(x\leq 5)=0.6009[/tex]

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