Mathematics, 15.07.2020 03:01 Christalex9874

5)One in four people in the US own individual stocks. You randomly select 12 people and ask them if they own individual stocks. a) Find the probability that the number who say they own individual stock is greater than 10. b) Find the probability that the numbers who say they own individual stock is less than 10. c) Find the probability that the numbers who say they own individual stock is at least 10.

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Answer from: Rattley6223

a) 0.0002% probability that the number who say they own individual stock is greater than 10.

b) 99.9963% probability that the numbers who say they own individual stock is less than 10.

c) 0.0037% probability that the numbers who say they own individual stock is at least 10.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they own individuals stocks, or they do not. The probability of a person owning stocks is independent of any other person. 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.

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

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.

One in four people in the US own individual stocks.

This means that $p = \frac{1}{4} = 0.25$

You randomly select 12 people

This means that n = 12

a) Find the probability that the number who say they own individual stock is greater than 10.

P(X 10) = P(X = 11) + P(X = 12)

In which

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

$P(X = 11) = C_{12,11}.(0.25)^{11}.(0.75)^{1} = 0.000002$

$P(X = 12) = C_{12,12}.(0.25)^{12}.(0.75)^{0} \approx 0$

P(X 10) = P(X = 11) + P(X = 12) = 0.000002 + 0 = 0.000002

0.0002% probability that the number who say they own individual stock is greater than 10.

b) Find the probability that the numbers who say they own individual stock is less than 10.

Either 10 or more say thew own, or less than 10 say. The sum of the probabilities of these outcomes is 1. So

$P(X \geq 10) + P(X < 10) = 1$

We want P(X < 10)

$P(X < 10) = 1 - P(X \geq 10)$

In which

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = 10) = C_{12,10}.(0.25)^{10}.(0.75)^{2} = 0.000035$

$P(X = 11) = C_{12,11}.(0.25)^{11}.(0.75)^{1} = 0.000002$

$P(X = 12) = C_{12,12}.(0.25)^{12}.(0.75)^{0} \approx 0$

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) = 0.000035 + 0.000002 + 0 = 0.000037$

$P(X < 10) = 1 - P(X \geq 10) = 1 - 0.000037 = 0.999963$

99.9963% probability that the numbers who say they own individual stock is less than 10.

c) Find the probability that the numbers who say they own individual stock is at least 10.

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) = 0.000035 + 0.000002 + 0 = 0.000037$

0.0037% probability that the numbers who say they own individual stock is at least 10.

Answer from: Quest

answer: approximately 206 bones

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Answer from: Quest

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