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.
a. To find the probability that at least one driver never uses a cell phone while driving out of a sample of 10 drivers, we can use the complement rule. The complement of "at least one driver never uses a cell phone while driving" is "no drivers never use a cell phone while driving."
The probability that a driver never uses a cell phone while driving is 15%, so the probability that a driver does use a cell phone while driving is 1 - 0.15 = 0.85.
The probability that all 10 drivers use a cell phone while driving is 0.85^10 ≈ 0.1967.
Therefore, the probability that at least one driver never uses a cell phone while driving is 1 - 0.1967 ≈ 0.8033.
b. To find the approximate probability that at least 13% of the 1000 drivers never use a cell phone while driving using the normal approximation to the binomial distribution, we need to calculate the mean and standard deviation.
Mean (μ) = n * p = 1000 * 0.15 = 150
Standard deviation (σ) = sqrt(n * p * (1 - p)) = sqrt(1000 * 0.15 * (1 - 0.15)) ≈ 10.95
Now, we can use the normal distribution to find the probability. We will use the continuity correction, subtracting 0.5 from 13 (0.13 - 0.5) to get the z-value.
z = (x - μ) / σ = (0.13 - 0.15) / 10.95 ≈ -0.0182
Using a standard normal distribution table or a calculator, we can find the probability associated with the z-value of -0.0182. The probability is approximately 0.4920.
Therefore, the approximate probability that at least 13% of these drivers never use a cell phone while driving is 1 - 0.4920 ≈ 0.5080.
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 found using a geometric distribution.
The probability that a driver never uses a cell phone while driving is 15%, so the probability that a driver uses a cell phone while driving is 1 - 0.15 = 0.85.
The probability that the first two drivers checked never use a cell phone while driving is (0.15)^2 = 0.0225.
Therefore, the probability that the first driver who uses a cell phone while driving is the third driver checked is 0.85 * 0.0225 ≈ 0.0191.
d. The expected number of drivers who must be checked to find the first who never uses a cell phone while driving can be calculated using the expected value of a geometric distribution.
The expected value of a geometric distribution with probability p is given by E(x) = 1/p.
In this case, the probability that a driver never uses a cell phone while driving is 15%, so p = 0.15.
Therefore, the expected number of drivers who must be checked to find the first who never uses a cell phone while driving is 1/0.15 ≈ 6.67.
e. The expected number of drivers who must be checked to find the first driver who uses a cell phone while driving can be calculated in a similar way.
The probability that a driver never uses a cell phone while driving is 15%, so the probability that a driver uses a cell phone while driving is 1 - 0.15 = 0.85.
Therefore, the expected number of drivers who must be checked to find the first driver who uses a cell phone while driving is 1/0.85 ≈ 1.18.
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 calculate the expected number of drivers who must be questioned.
As calculated in part e, the expected number of drivers who must be checked is approximately 1.18.
Given that it costs $5 to question each driver, the expected cost is 1.18 * $5 ≈ $5.90.
To find the standard deviation, we need to use the formula for the standard deviation of a geometric distribution, which is given by σ = sqrt((1 - p) / p^2).
In this case, p = 0.15, so the standard deviation is sqrt((1 - 0.15) / 0.15^2) ≈ 0.20.
Therefore, the standard deviation of questioning up to and including the first driver who uses a cell phone while driving is approximately $0.20 * $5 = $1.00.
g. The cost of inspection in part f (expected cost of $5.90) is not likely to exceed $15. Since the expected number of drivers who must be checked to find the first driver who uses a cell phone while driving is only 1.18, the cost of inspection is expected to be much lower than $15.
The expected cost of $5.90 is based on the probability of questioning all drivers up to and including the one who uses a cell phone while driving. It is unlikely to reach a cost of $15 in this scenario.