Question

According to a recent Census Bureau report, 12.7% of Americans live below the poverty level. Suppose you plan to sample at random 100 Americans and count the number of people who live below the poverty level. a. What is the probability that you count exactly 10 in poverty? b. What is the probability that you start taking the random sample and you find

Answer 1

To calculate the probabilities, we can use the binomial probability formula. For part (a), the probability of counting exactly 10 people below the poverty level can be found by substituting the values into the formula. For part (b), the question is incomplete, so a specific answer cannot be provided.

Step-by-step explanation:

To calculate the probabilities, we can use the binomial probability formula:

P(X=k) = C(n,k) * p^k * q^(n-k)

where:

- P(X=k) is the probability of getting exactly k successes

- C(n,k) is the number of ways to choose k items from a set of n items (combination)

- p is the probability of success

- q is the probability of failure (1-p)

In this case, we're interested in finding the probability of counting exactly 10 people living below the poverty level from a sample of 100 Americans, assuming the poverty rate is 12.7%.

To find the probability that exactly 10 people live below the poverty level, we have:

P(X=10) = C(100,10) * (0.127)^10 * (1-0.127)^(100-10)

Using a calculator or combinatorial calculator, we can find that C(100,10) = 17310309456440.

Substituting the values, we have:

P(X=10) = 17310309456440 * (0.127)^10 * (0.873)^90

Calculating this expression gives us the probability of counting exactly 10 people in poverty.

The question appears to be incomplete as it ends with 'you find'. Please provide the complete question for a more accurate answer.

Answer 2

a)
`P(X=10)=(100C10)(0.127)^(10) (1-0.127)^(100-10)=0.0928`

b)
`P(X \leq 10) = 0.2614`

c)
`(1-0.127)^7 (0.127) =0.0491`

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=100, p=0.127)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

a. What is the probability that you count exactly 10 in poverty?

For this case we want this probability P(X=10)

`P(X=10)=(100C10)(0.127)^(10) (1-0.127)^(100-10)=0.0928`

b. What is the probability that you count 10 or less in poverty? .2614

For this case we want this probability
`P(X=\leq10)`

`P(X\leq10)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)`

And we can find the individual probabilities like this:

`P(X=0)=(100C0)(0.127)^(0) (1-0.127)^(100-0)=1.263x10^(-6)`

`P(X=1)=(100C1)(0.127)^(1) (1-0.127)^(100-1)=1.837x10^(-5)`

`P(X=2)=(100C2)(0.127)^(2) (1-0.127)^(100-2)=0.000132`

`P(X=3)=(100C3)(0.127)^(3) (1-0.127)^(100-3)=0.000629`

`P(X=4)=(100C4)(0.127)^(4) (1-0.127)^(100-4)=0.00222`

`P(X=5)=(100C5)(0.127)^(5) (1-0.127)^(100-5)=0.00620`

`P(X=6)=(100C6)(0.127)^(6) (1-0.127)^(100-6)=0.0143`

`P(X=7)=(100C7)(0.127)^(7) (1-0.127)^(100-7)=0.0279`

`P(X=8)=(100C8)(0.127)^(8) (1-0.127)^(100-8)=0.0471`

`P(X=9)=(100C9)(0.127)^(9) (1-0.127)^(100-9)=0.0701`

`P(X=10)=(100C10)(0.127)^(10) (1-0.127)^(100-10)=0.0928`

And then repplacing we got:

`P(X \leq 10) = 0.2614`

c. What is the probability that you start taking the random sample and you find the first person in poverty on the 8th person selected? .0491

For this case we need after 7 people , 1 in poverty so we can find this probability like this:

`(1-0.127)^7 (0.127) =0.0491`