Normal Distribution Of Oil Change In A Car: Current Paramete

Normal Distributionchanging Oil In A Carcurrent Parameters Of Process

Normal Distributionchanging Oil In A Carcurrent Parameters Of Process

After studying the process of changing oil in a car, the shop’s manager has found that the distribution of service times, X, is normal with a mean of 28 minutes and a standard deviation of 5 minutes. Using this information, answer the following questions:

  1. What proportion of cars will be finished in less than half an hour (30 minutes)?
  2. What is the probability that a randomly selected car will take longer than 40 minutes?
  3. What service time corresponds to the 90th percentile?
  4. The manager wants to service 80 percent of the vehicles within 30 minutes. What must the mean service time be to accomplish this goal?

Use the z-table, NORMDIST, and NORMINV functions in Excel to determine the probabilities and critical values. Explain the steps involved and how the functions are applied.

Paper For Above instruction

The process of maintaining and servicing vehicles, specifically the time taken to change oil, is crucial for operational efficiency in automotive service shops. Understanding the distribution of service times helps optimize scheduling, resource allocation, and customer satisfaction. In this context, the service times for oil changes have been modeled as a normal distribution with a mean of 28 minutes and a standard deviation of 5 minutes, based on empirical observations. The following analysis explores various probabilistic measures and adjusts procedures to meet operational goals.

Calculating the Proportion of Cars Serviced in Less than 30 Minutes

Given that the service time, X, follows a normal distribution with mean μ = 28 minutes and standard deviation σ = 5 minutes, the first task is to determine the probability that a car is serviced in less than 30 minutes. This corresponds to calculating P(X

To compute this, we convert the raw score (30 minutes) to a z-score using the formula:

z = (X - μ) / σ = (30 - 28) / 5 = 0.4

Referring to the z-table, the area to the left of z = 0.4 is approximately 0.6554. Alternatively, in Excel, this probability can be found using the NORMDIST function:

=NORMDIST(30, 28, 5, TRUE)

This yields a probability of approximately 0.6554, indicating that about 65.54% of cars will be serviced in less than 30 minutes.

Probability of Taking Longer Than 40 Minutes

Next, we evaluate the likelihood that a vehicle exceeds 40 minutes of service, i.e., P(X > 40). First, compute the z-score:

z = (40 - 28) / 5 = 12 / 5 = 2.4

From the z-table, the area to the left of z = 2.4 is approximately 0.9918. Therefore, the probability of exceeding 40 minutes is:

P(X > 40) = 1 - 0.9918 = 0.0082

In Excel, this probability can be obtained using:

=1 - NORMDIST(40, 28, 5, TRUE)

This results in a probability of approximately 0.0082, or 0.82%, indicating a low chance of taking longer than 40 minutes.

Determining the 90th Percentile Service Time

The 90th percentile corresponds to the service time below which 90% of the observations fall. We find the z-score corresponding to the 90th percentile using the inverse cumulative distribution function (NORMINV in Excel):

In Excel:

=NORMINV(0.9, 28, 5)

This yields a z-score of approximately 1.2816. Converting this z-score back to the actual service time:

Service time = μ + z σ = 28 + 1.2816 5 ≈ 28 + 6.408 = 34.408 minutes

Thus, approximately 90% of oil changes are completed in less than 34.41 minutes.

Adjusting the Mean Service Time to Service 80% of Vehicles within 30 Minutes

The shop manager aims to service 80% of vehicles within 30 minutes, which means P(X

In Excel:

=NORMINV(0.8, 0, 1)

This gives a z-score of approximately 0.8416. Now, solving for μ:

μ = X - z σ = 30 - 0.8416 5 ≈ 30 - 4.208 = 25.792 minutes

Therefore, to achieve the goal of servicing 80% of cars within 30 minutes, the average service time should be reduced to approximately 25.79 minutes.

Conclusion

Analyzing service times through the normal distribution provides valuable insights into operational performance and targets. The calculated probabilities assist managers in understanding variability and setting realistic goals. By utilizing statistical functions such as NORMDIST and NORMINV, service shops can optimize schedules, improve efficiency, and increase customer satisfaction. The ability to adjust the mean service time based on desired throughput exemplifies how statistical analysis directly informs operational strategies.

References

  • Ohri, D. (2013). Statistics for Business and Economics. Cengage Learning.
  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
  • Mendenhall, W., Beaver, R., & Beaver, B. (2013). Introduction to Probability and Statistics. Thomson Brooks/Cole.
  • Montgomery, D. C., & Runger, G. C. (2014). Probability and Statistics for Engineering and the Sciences. Wiley.
  • Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability & Statistics for Engineering and the Sciences. Pearson.
  • Freund, J. E., & Walpole, R. E. (2014). Mathematical Statistics with Applications. Pearson.
  • Hogg, R. V., & Tanis, E. A. (2013). Probability and Statistical Inference. Pearson.
  • Zwillinger, D. (2018). Table of Integrals, Series, and Products. Academic Press.
  • Excel Documentation. (2020). Functions: NORMDIST, NORMINV. Microsoft.
  • Statistics How To. (2021). Normal Distribution and Z-Score Calculations.

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