Suppose that you are studying the number of vehicles owned by a US household. You find that the average is 1.8 vehicles per household with a standard deviation of 2.9 vehicles per household.

(a) What is the most likely shape for the population described above?
(b) Find the probability that 45 randomly selected households have an AVERAGE number of vehicles GREATER than 2.5.
(c) Find the probability that 75 randomly selected households have a TOTAL number of vehicles LESS than 150.

1 answer

(a) The most likely shape for the population described would be a normal distribution. This is because averages of a large number of independent and identically distributed random variables tend to follow a normal distribution due to the Central Limit Theorem.

(b) To find the probability that 45 randomly selected households have an average number of vehicles greater than 2.5, we need to calculate the z-score and the corresponding probability.

First, we need to calculate the standard error of the mean (SEM), which is equal to the standard deviation divided by the square root of the sample size:

SEM = 2.9 / √45 ≈ 0.432

Next, we can calculate the z-score using the formula:

z = (sample mean - population mean) / SEM

In this case, the population mean is 1.8 and the sample mean we want to compare is 2.5:

z = (2.5 - 1.8) / 0.432 ≈ 1.620

Using a standard normal distribution table or a calculator, we can find the probability corresponding to this z-score. The probability is approximately 0.0538, or 5.38%.

Therefore, the probability that 45 randomly selected households have an average number of vehicles greater than 2.5 is approximately 5.38%.

(c) To find the probability that 75 randomly selected households have a total number of vehicles less than 150, we need to calculate the z-score and the corresponding probability.

First, we need to calculate the mean and standard deviation of the total number of vehicles. Since the average number of vehicles per household is 1.8 and the standard deviation is 2.9, the mean and standard deviation of the total number of vehicles can be calculated as follows:

Mean of total number of vehicles = 1.8 * 75 = 135
Standard deviation of total number of vehicles = 2.9 * √75 ≈ 20.15

Next, we need to calculate the z-score using the formula:

z = (sample total - population mean) / (standard deviation / √sample size)

In this case, the population mean is 135 and the sample total we want to compare is 150:

z = (150 - 135) / (20.15 / √75) ≈ 1.06

Using a standard normal distribution table or a calculator, we can find the probability corresponding to this z-score. The probability is approximately 0.8577, or 85.77%.

Therefore, the probability that 75 randomly selected households have a total number of vehicles less than 150 is approximately 85.77%.