91Ó°ÊÓ

The shape of the distribution of the time required to get an oil change at a 10 -minute oil-change facility is unknown. However, records indicate that the mean time for an oil change is 11.4 minutes, and the standard deviation for oilchange time is 3.2 minutes. (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? (b) What is the probability that a random sample of \(n=40\) oil changes results in a sample mean time of less than 10 minutes? (c) Suppose the manager agrees to pay each employee a \(\$ 50\) bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, what mean oil-change time would there be a \(10 \%\) chance of being at or below? This will be the goal established by the manager.

Step by step solution

01

Determine sample size for normal model

To use the normal model for the sample mean, the Central Limit Theorem states that the sample size () should be at least 30 for the distribution of the sample mean to be approximately normal, regardless of the shape of the population distribution. Therefore, a sample size of 30 or more is required.

02

Calculate the standard error of the mean

The standard error of the mean (SEM) is given by the formula: \( \text{SEM} = \frac{\text{standard deviation}}{\text{sqrt}(n)} \) Given that the standard deviation (σ) is 3.2 minutes and the sample size (n) is 40: \( \text{SEM} = \frac{3.2}{\text{sqrt}(40)} \) \( \text{SEM} \approx 0.506 \)

03

Calculate the z-score for a sample mean of 10 minutes

The z-score is calculated using the formula: \( z = \frac{\text{sample mean} - \text{population mean}}{\text{SEM}} \) Substitute the values provided: \( z = \frac{10 - 11.4}{0.506} \) \( z \approx -2.77 \)

04

Determine the probability of the z-score

Using the standard normal distribution table, find the probability corresponding to a z-score of -2.77. This probability is approximately 0.0028. Therefore, the probability that a random sample of n=40 oil changes results in a sample mean time of less than 10 minutes is approximately 0.0028 or 0.28%.

05

Calculate z-score for the required goal (10% chance)

To set a goal with a 10% chance of being met, find the z-score that corresponds to the 10th percentile in the standard normal distribution, which is approximately -1.28.

06

Calculate the corresponding sample mean time for the goal

Using the z-score formula, rearrange to find the sample mean (\( \bar{x} \)): \( z = \frac{\bar{x} - \text{population mean}}{\text{SEM}} \) Solve for \( \bar{x} \): \( \bar{x} = z \times \text{SEM} + \text{population mean} \) \( \bar{x} = -1.28 \times 0.506 + 11.4 \) \( \bar{x} \approx 10.75 \) Thus, the goal set by the manager would be a mean oil-change time of 10.75 minutes.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution

The normal distribution is a continuous probability distribution that is symmetrical around its mean. When graphed, it forms a bell-shaped curve that depicts the spread of a dataset. The key characteristics are that about 68% of data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations. In this oil-change scenario, we use the normal distribution to determine probabilities related to the sample mean. Using the Central Limit Theorem, we know that if the sample size is large enough (typically n > 30), the distribution of the sample mean will approximate a normal distribution, even if the original data is not normally distributed.

Sample Mean

The sample mean is the average value of a sample, and it serves as an estimate of the population mean. In a study where we want to calculate how long an oil change takes on average, the sample mean represents the average time from our collected data. For example, if you record 40 different oil changes and find their average, that is your sample mean. The value of the sample mean will vary from sample to sample but can give us insight into the population mean, which is the true average time it takes for an oil change. In this exercise, the sample mean helps us establish a threshold whereby employees get a bonus if the mean service time is below a particular value.

Standard Error

The standard error (SE) quantifies the extent of variable fluctuation from the sample mean to the population mean. It gives us insight into how accurately our sample mean represents the population mean. The formula for the standard error of the mean (SEM) is: \( SEM = \frac{\sigma}{\sqrt{n}} \) where \( \sigma \) is the population standard deviation and \( n \) is the sample size. In our example, given \( \sigma = 3.2 \) and a sample size \( n = 40\), the SEM is calculated as: \( SEM = \frac{3.2}{\sqrt{40}} \) which approximates to 0.506. A smaller SEM indicates a more precise estimate of the population mean. Understanding the standard error is crucial when we use the normal distribution to determine probabilities for the sample mean.