A survey of drivers in the United States found that 15% never use a cell phone while driving. Suppose that drivers arrive at random at an auto inspection station.
a. If the inspector checks 10 drivers, what is the probability that at least one driver never uses a cell phone while driving?
b. Suppose the inspector checks 1000 drivers. Use the normal approximation to the binomial distribution to find the approximate probability that at least 13% of these drivers never use a cell phone while driving.
c. If the drivers are inspected sequentially as they arrive randomly at the inspection station, what is the probability that the first driver who uses a cell phone while driving is the third driver checked?
d. What is the expected number of drivers who must be checked to find the first who never uses a cell phone while driving?
e. What is the expected number of drivers who must be checked to find the first driver who uses a cell phone while driving?
f. If it costs $5 to question each driver, what is the expected cost and standard deviation of questioning up to and including the first driver who uses a cell phone while driving?
g. Will the cost of inspection in part f often exceed $15? Explain.
I'm not sure..
a. To find the probability that at least one driver never uses a cell phone while driving, you can use the complement rule.
The complement of "at least one driver never uses a cell phone while driving" is "no driver never uses a cell phone while driving".
Since we know that 15% of drivers never use a cell phone while driving, the probability that a driver does use a cell phone while driving is 1 - 0.15 = 0.85.
Now we can calculate the probability using the complement rule.
P(at least one driver never uses a cell phone while driving) = 1 - P(no driver never uses a cell phone while driving)
P(at least one driver never uses a cell phone while driving) = 1 - (0.85)^10
b. To use the normal approximation to the binomial distribution, we'll use the mean and standard deviation formula.
The mean (μ) of a binomial distribution is given by μ = n * p, where n is the number of trials and p is the probability of success.
The standard deviation (σ) of a binomial distribution is given by σ = √(n * p * (1 - p))
In this case, n = 1000 and p = 0.15.
Therefore, μ = 1000 * 0.15 = 150 and σ = √(1000 * 0.15 * (1 - 0.15)).
We want to find the approximate probability that at least 13% of these drivers never use a cell phone while driving, which is equivalent to finding the probability that X ≥ 0.13 * 1000 = 130.
Using the normal approximation for the binomial distribution, we can calculate P(X ≥ 130) using the Z-score formula.
Z = (X - μ) / σ
P(X ≥ 130) = P(Z ≥ (130 - μ) / σ)
c. If the drivers are inspected sequentially as they arrive randomly at the inspection station, the probability that the first driver who uses a cell phone while driving is the third driver checked can be calculated using the geometric distribution.
The probability of success (p) is 0.85, and we want to find the probability that it takes exactly 3 trials to get the success.
P(X = 3) = (1 - p)^2 * p
d. The expected number of drivers who must be checked to find the first who never uses a cell phone while driving is given by the formula for the expected value of a geometric distribution.
E(X) = 1 / p
In this case, p = 0.15.
e. The expected number of drivers who must be checked to find the first driver who uses a cell phone while driving is also given by the formula for the expected value of a geometric distribution.
E(X) = 1 / p
In this case, p = 0.85.
f. The expected cost of questioning up to and including the first driver who uses a cell phone while driving can be found by multiplying the expected number of drivers with the cost per driver.
Expected cost = E(X) * cost per driver = (1 / p) * cost per driver
g. Whether the cost of inspection in part f will often exceed $15 depends on the specific values of p (probability of success), cost per driver, and the rounding of the expected cost. Without those values, it is not possible to determine if the cost will often exceed $15.
To solve these problems, we will use the binomial distribution.
a. The probability of at least one driver never using a cell phone while driving in a sample of 10 drivers can be calculated as the complement of the probability that all drivers use a cell phone. The probability that a driver uses a cell phone while driving is 1 - 0.15 = 0.85.
P(at least one driver never uses a cell phone) = 1 - P(all drivers use a cell phone)
P(at least one driver never uses a cell phone) = 1 - (0.85)^10
P(at least one driver never uses a cell phone) ≈ 1 - 0.196 = 0.804 or 80.4%
b. To approximate the probability that at least 13% of the 1000 drivers never use a cell phone, we can calculate the mean and standard deviation of the binomial distribution and use the normal approximation.
Mean (μ) = n * p = 1000 * 0.15 = 150
Standard Deviation (σ) = sqrt(n * p * (1 - p)) = sqrt(1000 * 0.15 * 0.85) ≈ 9.91
Now, we need to find the probability that at least 13% (0.13) of the 1000 drivers never use a cell phone.
P(at least 13% never use a cell phone) ≈ P(x ≥ 0.13 * 1000) ≈ P(x ≥ 130)
To use the normal approximation, we need to standardize the value:
Z = (x - μ) / σ = (130 - 150) / 9.91 ≈ -2.02
Using a standard normal distribution table or a calculator, we can find the probability associated with this value of Z, and then subtract it from 1 to find the probability we need:
P(x ≥ 130) ≈ 1 - P(Z ≤ -2.02)
c. The probability that the first driver who uses a cell phone while driving is the third driver checked can be calculated as follows:
P(first driver who uses a phone is the third driver) = (0.85)^2 * (0.15) ≈ 0.108
d. The expected number of drivers who must be checked to find the first who never uses a cell phone while driving is the reciprocal of the probability of finding the first non-phone user.
Expected number of drivers = 1 / (0.15) = 6.67 (rounded to 2 decimal places)
e. The expected number of drivers who must be checked to find the first driver who uses a cell phone while driving is the reciprocal of the probability of finding the first phone user.
Expected number of drivers = 1 / (0.85) = 1.18 (rounded to 2 decimal places)
f. To find the expected cost and standard deviation of questioning up to and including the first driver who uses a cell phone while driving, we need to multiply the expected number of drivers from part e by the cost per driver.
Expected cost = Expected number of drivers * Cost per driver = 1.18 * $5 = $5.90 (rounded to 2 decimal places)
To calculate the standard deviation, we use the fact that the standard deviation of a binomial distribution is given by the formula:
Standard deviation (σ) = sqrt(n * p * (1 - p))
Standard deviation = sqrt(1 * 0.85 * 0.15) = 0.367
Standard deviation of cost = Standard deviation * Cost per driver = 0.367 * $5 ≈ $1.84
g. No, the cost of inspection in part f will not often exceed $15. Since the expected number of drivers needed to find the first phone user is 1.18, the expected cost will most often be less than $10, and the standard deviation of the cost is approximately $1.84. Thus, the cost is expected to be within one standard deviation of the mean ($5.90) in the vast majority of cases, making it highly unlikely for it to exceed $15.