it is estimated that 17% of americans have blue eyes. A random sample of 9 americans is selected. Find the probability that the sample includes exactly 2 people with blue eyes. Find the probability that the sample includes at most 2 people with blue eyes. Find the mean number of blu-eyed people in a sample of 9 americans.

Mean = np = 9 * .17 = 1.53

Use binomial probability formula (or a binomial probability table).

Formula:
P(x) = (nCx)(p^x)[q^(n-x)]

x = 0,1,2
n = 9
p = .17
q = 1 - p = 1 - .17 = .83

For first part: Find P(2) for probability.

For second part: Find P(0) and P(1). Add P(0), P(1), and P(2) for probability.

I'll let you take it from here.

To find the probability that a sample of 9 Americans includes exactly 2 people with blue eyes, we can use the binomial probability formula. The formula is:

P(X = k) = (n Choose k) * (p^k) * ((1-p)^(n-k))

where:
P(X = k) is the probability of getting exactly k successes,
(n Choose k) is the number of ways to choose k successes from n trials, given by the formula (n! / (k! * (n-k)!)),
p is the probability of success in a single trial,
k is the number of successes, and
n is the number of trials.

In this case, n (number of trials) is 9 and p (probability of success) is 0.17 (17% or 0.17 chance of having blue eyes). We want to find P(X = 2), so plugging in the values:

P(X = 2) = (9 Choose 2) * (0.17^2) * ((1-0.17)^(9-2))

Calculating this:

P(X = 2) = (9! / (2! * (9-2)!)) * (0.17^2) * (0.83^7)

P(X = 2) = (36/2) * (0.0289) * (0.243)

P(X = 2) ≈ 0.098

Therefore, the probability that the sample includes exactly 2 people with blue eyes is approximately 0.098, or 9.8%.

To find the probability that the sample includes at most 2 people with blue eyes, we need to find the cumulative probability. This means we need to calculate the probabilities for k = 0, 1, and 2, and then add them together. Let's calculate each probability separately and then sum them:

P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2)

To find P(X = 0), we can use the same formula as before:

P(X = 0) = (9 Choose 0) * (0.17^0) * ((1-0.17)^(9-0))

P(X = 0) = (9! / (0! * (9-0)!)) * (1) * (0.83^9)

P(X = 0) = (1) * (1) * (0.83^9)

P(X = 0) ≈ 0.343

To find P(X = 1) and P(X = 2), we follow the same process as before. Plugging in the values:

P(X = 1) ≈ 0.391
P(X = 2) ≈ 0.098

Now we can sum these probabilities:

P(X <= 2) = 0.343 + 0.391 + 0.098

P(X <= 2) ≈ 0.832

Therefore, the probability that the sample includes at most 2 people with blue eyes is approximately 0.832, or 83.2%.

To find the mean number of blue-eyed people in a sample of 9 Americans, we can use the expected value formula. The formula is:

E(X) = n * p

where:
E(X) is the expected value (mean),
n is the number of trials, and
p is the probability of success in a single trial.

In this case, n (number of trials) is 9 and p (probability of success) is 0.17 (17% or 0.17 chance of having blue eyes). Plugging in the values:

E(X) = 9 * 0.17

E(X) = 1.53

Therefore, the mean number of blue-eyed people in a sample of 9 Americans is approximately 1.53.