The average price of a gallon of unleaded regular gasoline was reported to be $2.34 in northern Kentucky (The Cincinnati Enquirer, January 21, 2006). Use this price as the population mean, and assume the population standard deviation is $.20.
1. What is the probability that the mean price for a sample of 30 service stations is within $.03 of the population mean (to 4 decimals)?
2. What is the probability that the mean price for a sample of 50 service stations is within $.03 of the population mean (to 4 decimals)?
3. What is the probability that the mean price for a sample of 100 service stations is within $.03 of the population mean (to 4 decimals)?
4. Calculate the sample size necessary to guarantee at least .95 probability that the sample mean is within $.03 of the population mean (0 decimals).
To solve these questions, we need to use the properties of the normal distribution. The probability distribution of the sample mean is normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
1. To find the probability that the mean price for a sample of 30 service stations is within $0.03 of the population mean, we can compute the z-score for this range and then use a standard normal distribution table or calculator to find the probability.
First, we calculate the standard error of the mean (SE):
SE = population standard deviation / sqrt(sample size)
SE = $0.20 / sqrt(30)
Next, we calculate the z-score:
z = (sample mean - population mean) / SE
z = ($2.34 - $2.34) / ($0.20 / sqrt(30))
Since the range is within $0.03 of the population mean, we can calculate the probability using the cumulative distribution function (CDF) of the standard normal distribution or a standard normal distribution table.
2. We use the same method as in question 1, but with the sample size of 50.
SE = $0.20 / sqrt(50)
z = ($2.34 - $2.34) / ($0.20 / sqrt(50))
3. Again, we use the same method as in question 1, but with the sample size of 100.
SE = $0.20 / sqrt(100)
z = ($2.34 - $2.34) / ($0.20 / sqrt(100))
4. To calculate the sample size necessary to guarantee at least 0.95 probability that the sample mean is within $0.03 of the population mean, we need to find the z-score associated with the desired confidence level. We can then rearrange the formula for the z-score calculation to solve for the required sample size.
Using the inverse of the cumulative distribution function (CDF) of the standard normal distribution for a given confidence level, we find the z-score. From there, we rearrange the formula for the z-score:
z = ($0.03) / (population standard deviation / sqrt(sample size))
Solving for sample size:
sqrt(sample size) = ($0.03) / (population standard deviation * z)
sample size = [($0.03) / (population standard deviation * z)]^2
You can substitute the values ($0.03, population standard deviation, and desired confidence level z-score) into the above formula to calculate the required sample size.
Remember to round the probabilities to 4 decimal places and the sample size to 0 decimals.