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According to a Gallup poll, 60% of American adults prefer saving over spending. In a random sample of 10 American adults, what is the probability that more than 3 adults in the sample prefer saving over spending

Sagot :

Answer:

0.9452 = 94.52% probability that more than 3 adults in the sample prefer saving over spending

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they prefer saving over spending, or they do not. The answers for each adult are independent, which 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.

[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.

According to a Gallup poll, 60% of American adults prefer saving over spending.

This means that [tex]p = 0.6[/tex]

Sample of 10 American adults

This means that [tex]n = 10[/tex]

What is the probability that more than 3 adults in the sample prefer saving over spending?

This is:

[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]

In which

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{10,0}.(0.6)^{0}.(0.4)^{10} = 0.0001[/tex]

[tex]P(X = 1) = C_{10,1}.(0.6)^{1}.(0.4)^{9} = 0.0016[/tex]

[tex]P(X = 2) = C_{10,2}.(0.6)^{2}.(0.4)^{8} = 0.0106[/tex]

[tex]P(X = 3) = C_{10,3}.(0.6)^{3}.(0.4)^{7} = 0.0425[/tex]

Then

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0001 + 0.0016 + 0.0106 + 0.0425 = 0.0548[/tex]

[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.0548 = 0.9452[/tex]

0.9452 = 94.52% probability that more than 3 adults in the sample prefer saving over spending

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