According to the CDC, 25 % of college students get the flu every year.

(a) If 7 college students are selected at random? What is the probability that exactly 6 of them have the flu?
(b) What is the mean and standard deviation for this distribution?

Hey, this is just another binomial dist prob

try
http://ncalculators.com/statistics/binomial-distribution-calculator.htm

an insurance policy sells for $700. based on past data, an average of 1 in 80 policyholders will file a $10,000 claim, an average of 1 in 120 policyholders will file a $20,000 claim, and an average of 1 in 200 policyholders will file a $50,000 claim. what is the expected value the company per policy sold?

To find the probability that exactly 6 of the 7 randomly selected college students have the flu, we can use the binomial probability formula. The formula for the probability of exactly k successes (having the flu) in n trials (number of students selected) is:

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

Where:
- C(n, k) represents the number of ways to choose k successes out of n trials, also known as the binomial coefficient. It can be calculated as C(n, k) = n! / (k! * (n - k)!)
- p is the probability of a single success (getting the flu), which is given as 25%, so we write it as 0.25 (or 0.25/1).

(a) Probability calculation:
In this case, n (number of trials) is 7, k (number of successes) is 6, and p (probability of a single success) is 0.25.

P(X = 6) = C(7, 6) * 0.25^6 * (1 - 0.25)^(7 - 6)

To calculate C(7, 6):
C(7, 6) = 7! / (6! * (7 - 6)!)
= 7! / (6! * 1!)
= 7

Substituting the values into the probability formula:
P(X = 6) = 7 * 0.25^6 * 0.75^1

Calculating the probability:
P(X = 6) = 7 * (0.25)^6 * (0.75)^1
= 0.007172

Therefore, the probability that exactly 6 out of 7 randomly selected college students have the flu is approximately 0.007172.

(b) Mean and standard deviation calculation:
The mean (μ) and standard deviation (σ) for a binomial distribution can be calculated as follows:

Mean (μ) = n * p
Standard Deviation (σ) = √(n * p * (1 - p))

In this case, n (number of trials) is 7 and p (probability of a single success) is 0.25.

Mean (μ) = 7 * 0.25 = 1.75
Standard Deviation (σ) = √(7 * 0.25 * 0.75) = √(1.3125) = 1.145

Therefore, the mean (μ) for this distribution is 1.75, and the standard deviation (σ) is approximately 1.145.