Answer:
a. To solve this problem, we can use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- X is the number of out-of-state vehicles in a sample of n vehicles
- k is the number of out-of-state vehicles we're interested in (in this case, k=3)
- p is the probability that a given vehicle is from out of state (in this case, p=0.25)
- n is the sample size (in this case, n=9)
Plugging in the values, we get:
P(X=3) = (9 choose 3) * 0.25^3 * 0.75^6
= 84 * 0.0039 * 0.1785
= 0.5907%
Therefore, the probability that exactly three of the next 9 vehicles are from out of state is 0.5907%.
b. To find the expected number of out-of-state vehicles in the next hour, we can use the formula:
E(X) = n * p
where:
- X is the number of out-of-state vehicles in a sample of n vehicles
- p is the probability that a given vehicle is from out of state (in this case, p=0.25)
- n is the sample size (in this case, n=140)
Plugging in the values, we get:
E(X) = 140 * 0.25
= 35
Therefore, the expected number of vehicles from out of state in the next hour is 35.
c. To find the probability of the number of vehicles varying between 2 standard deviations from the mean number of those passing through the check point, we need to find the mean and standard deviation of the number of out-of-state vehicles in a sample of 140 vehicles.
The mean is simply the expected value we found in part b:
mean = 35
The variance of a binomial distribution is:
Var(X) = n * p * (1-p)
where:
- X is the number of out-of-state vehicles in a sample of n vehicles
- p is the probability that a given vehicle is from out of state (in this case, p=0.25)
- n is the sample size (in this case, n=140)
Plugging in the values, we get:
Var(X) = 140 *
hope this helps :o