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.