Suppose cars arrive at Burger King's drive-through at the rate of 20 cars every hour between 12:00 noon and 1:00 pm. A random sample of 40 one-hour periods between 12:00 noon and 1:00 pm is selected and has 22.1 as the sample mean number of cars arriving. (a) Why is the sampling distribution of the sample mean approximately normal? (b) Assuming and , what is the mean and standard deviation of the sampling distribution? What is the probability that a simple random sample of size n = 40 one-hour time periods results in a mean of at least 22.1 cars? Is this result unusual?
a. n is a large number
b. mean = 12 sd = 4
c. 0.45
(a) The sampling distribution of the sample mean is approximately normal due to the Central Limit Theorem. According to the Central Limit Theorem, when the sample size is sufficiently large, the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution. In this case, we have a random sample of 40 one-hour periods, which can be considered sufficiently large for the Central Limit Theorem to apply.
(b) Given that the population mean arrival rate is 20 cars per hour, denoted as μ, and the population standard deviation is σ, we need to find the mean and standard deviation of the sampling distribution of the sample mean.
The mean of the sampling distribution, denoted as μₙ, is equal to the population mean (μ). Therefore, μₙ = μ = 20 cars per hour.
The standard deviation of the sampling distribution, denoted as σₙ, is equal to the population standard deviation (σ) divided by the square root of the sample size (n). Therefore, σₙ = σ / sqrt(n).
Since we are not given the population standard deviation (σ) in this problem, we cannot calculate the exact standard deviation of the sampling distribution. However, we can make an estimation using the sample standard deviation (s) as a substitute for σ. This is common when the population standard deviation is unknown.
To calculate the probability that a simple random sample of size n = 40 one-hour time periods results in a mean of at least 22.1 cars, we can use the Z-score formula.
Z = (x - μₙ) / (σₙ)
In this case, x = 22.1, μₙ = 20, and we can estimate σₙ using the sample standard deviation.
First, we need to estimate the population standard deviation (σ) using the sample standard deviation (s). Assuming the sample standard deviation is provided, we can substitute it into the formula:
σₙ ≈ s / sqrt(n)
σₙ ≈ (sample standard deviation) / sqrt(n)
Once we have an estimate for σₙ, we can calculate the Z-score:
Z = (x - μₙ) / σₙ
Since we do not have the sample standard deviation (s) provided in the question, we cannot complete the calculations and provide the exact probability. However, once the sample standard deviation is known, we can look up the Z-score in a standard normal distribution table or use a statistical calculator to find the probability associated with the Z-score. A higher Z-score would indicate a lower probability.
Whether the result is unusual or not depends on the significance level chosen. If the probability is below the chosen significance level (e.g., 0.05), it is considered unusual. Otherwise, it is considered not unusual.