To solve this problem, we will use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Step 1: Identify the given values.
The population mean (μ) = 114114 gallons
The standard deviation (σ) = 77 gallons
The sample size (n) = 4646 cars
The difference from the mean we are interested in (d) = 11 gallons

Step 2: Calculate the standard error (SE), which is the standard deviation of the sampling distribution. The formula for standard error is σ/√n.
SE = 77/√4646 = 1.131 gallons

Step 3: Calculate the z-score, which is the number of standard errors a particular sample mean is from the population mean. The formula for the z-score is (X - μ)/SE, where X is the sample mean. In this case, we are looking for the probability that the sample mean differs from the population mean by more than 11 gallons, so X = μ ± d = 114114 ± 11 = 114125 and 114103.

Z1 = (114125 - 114114)/1.131 = 9.72
Z2 = (114103 - 114114)/1.131 = -9.72

Step 4: Look up the z-scores in the z-table to find the probabilities. The z-table shows the probability that the sample mean is less than the population mean by a certain number of standard errors (the z-score). So, to find the probability that the sample mean is greater than the population mean by 9.72 standard errors, we need to find the probability that the z-score is less than -9.72, which is almost 0. Similarly, the probability that the z-score is less than 9.72 is almost 1.

Step 5: Subtract the two probabilities to find the probability that the sample mean differs from the population mean by more than 11 gallons.
P = P(Z

Question

To solve this problem, we will use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Step 1: Identify the given values.
The population mean (μ) = 114114 gallons
The standard deviation (σ) = 77 gallons
The sample size (n) = 4646 cars
The difference from the mean we are interested in (d) = 11 gallons

Step 2: Calculate the standard error (SE), which is the standard deviation of the sampling distribution. The formula for standard error is σ/√n. 
SE = 77/√4646 = 1.131 gallons

Step 3: Calculate the z-score, which is the number of standard errors a particular sample mean is from the population mean. The formula for the z-score is (X - μ)/SE, where X is the sample mean. In this case, we are looking for the probability that the sample mean differs from the population mean by more than 11 gallons, so X = μ ± d = 114114 ± 11 = 114125 and 114103.

Z1 = (114125 - 114114)/1.131 = 9.72
Z2 = (114103 - 114114)/1.131 = -9.72

Step 4: Look up the z-scores in the z-table to find the probabilities. The z-table shows the probability that the sample mean is less than the population mean by a certain number of standard errors (the z-score). So, to find the probability that the sample mean is greater than the population mean by 9.72 standard errors, we need to find the probability that the z-score is less than -9.72, which is almost 0. Similarly, the probability that the z-score is less than 9.72 is almost 1.

Step 5: Subtract the two probabilities to find the probability that the sample mean differs from the population mean by more than 11 gallons.
P = P(Z < 9.72) - P(Z < -9.72) = 1 - 0 = 1

So, the probability that the sample mean would differ from the population mean by greater than 11 gallons is almost 1, or 100%. This means it is almost certain that the sample mean will differ from the population mean by more than 11 gallons.

Please note that the values of the mean and standard deviation seem to be incorrect or unusually high, which may have affected the result.

Knowee AI · Accepted Answer