An automobile repair shop has determined that the average service time on an automobile is 2 hours with a standard deviation of 32 minutes. A random sample of 64 out of 400 services is selected. What is the probability that this sample will have a mean service time greater than 114 minutes?

.06

0.6

To find the probability that the sample mean service time is greater than 114 minutes, we need to use the Central Limit Theorem.

According to the Central Limit Theorem, the sampling distribution of the sample mean approaches a normal distribution as the sample size increases.

The mean of the sampling distribution of the sample mean is equal to the population mean, which is 2 hours or 120 minutes in this case.

The standard deviation of the sampling distribution of the sample mean, also known as the standard error, can be calculated using the formula:

Standard Error = Population Standard Deviation / √(Sample Size)

In this case, the population standard deviation is 32 minutes, and the sample size is 64.

Standard Error = 32 / √(64) = 32 / 8 = 4

Now, we need to calculate the z-score for a sample mean of 114 minutes using the formula:

z = (x - μ) / σ

Where:
x = sample mean (114 minutes)
μ = population mean (120 minutes)
σ = standard error (4 minutes)

z = (114 - 120) / 4 = -6 / 4 = -1.5

To find the probability that the sample mean is greater than 114 minutes, we need to find the area to the right of the z-score of -1.5 on a standard normal distribution.

Using a standard normal distribution table or calculator, we can find that the probability corresponding to a z-score of -1.5 is approximately 0.0668.

However, since we are interested in the probability that the sample mean is greater than 114 minutes, we need to find the area to the right of -1.5, which is 1 - 0.0668 = 0.9332.

Therefore, the probability that the sample will have a mean service time greater than 114 minutes is approximately 0.9332 or 93.32%.

To solve this problem, we need to use the central limit theorem and the properties of the sample mean.

The central limit theorem states that the distribution of sample means from a population with any shape will become approximately normal as the sample size increases. In this case, we have a large enough sample size (64), so we can assume the sample mean follows a normal distribution.

To calculate the probability that the sample mean is greater than 114 minutes, we need to standardize the sample mean using the formula for the z-score:

z = (x - μ) / (σ / sqrt(n))

Where:
- x is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size

In this case:
- x = 114 minutes
- μ = 120 minutes (2 hours)
- σ = 32 minutes
- n = 64

Plugging in these values to the formula, we get:

z = (114 - 120) / (32 / sqrt(64))
= -6 / (32 / 8)
= -6 / 4
= -1.5

Now, we need to find the probability that the z-score is greater than -1.5. We can use a standard normal distribution table or a calculator to find this probability.

Using a standard normal distribution table, we can find the probability corresponding to a z-score of -1.5. The probability for a z-score greater than -1.5 is approximately 0.9332. Therefore, the probability that the sample will have a mean service time greater than 114 minutes is 0.9332, or 93.32%.