Question

A traffic control engineer reports that 75% of the vehicles passing through a check poin from within the state. [15 points]

a. What is the probability that exactly three of the next 9 vehicles are from out of state? 5 points
b. It is estimate that 140 vehicles will go through this check point in the next hour What is the expected number of vehicles from out of state in the next hour? 5
c. What is the probability of the number of vehicles varying between 2 standard deviations from the mean number of those passing through the check point

Answer 1

a. To find the probability that exactly 3 of the next 9 vehicles are from out of state, we can use the binomial probability formula:

P(X=3) = (9 choose 3) * (0.25)^3 * (0.75)^6 = 0.250

Therefore, the probability that exactly three of the next 9 vehicles are from out of state is 0.250.

b. To find the expected number of vehicles from out of state in the next hour, we can use the formula for the mean of a binomial distribution:

E(X) = n*p = 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 standard deviation of the binomial distribution:

σ = sqrt(n*p*(1-p)) = sqrt(140*0.25*0.75) = 5.369

The mean number of vehicles passing through the checkpoint is 35, so 2 standard deviations from the mean is 35 ± 2*5.369 = (24.262, 45.738).

To find the probability of the number of vehicles varying between 2 standard deviations from the mean, we can use the normal approximation to the binomial distribution:

P(24.262 < X < 45.738) = P(Z < (45.738-35)/5.369) - P(Z < (24.262-35)/5.369) = P(Z < 1.867) - P(Z < -2.042) = 0.968 - 0.020 = 0.948

Therefore, the probability of the number of vehicles varying between 2 standard deviations from the mean is 0.948.

Answer 2

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