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Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a \(50 \%\) discount on the charges. The company wants to limit this discount to at most \(5 \%\) of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.

Short Answer

Expert verified
The maximum guaranteed wait time should be 19 minutes.

Step by step solution

01

Understanding the problem and creating a graphical representation

According to the problem, all the times taken for service at Fast Auto Service follow a Normal distribution with a mean of 15 and a standard deviation of 2.4. The garage is planning to give a 50% discount to customers who have to wait longer than the guaranteed waiting time. The garage wants less than or equal to 5% of the customers to get this discount. This means that 95% of the customers' serving time should be less than or equal to the guaranteed waiting time, according to the properties of Normal distribution.
02

Calculate the Z-Score

The Z-score for 0.95 (or 95%) can be obtained from the standard Normal distribution table or using a statistical computing software, which gives us approximately 1.645. Remember that the z-score gives us the number of standard deviations a particular value is from the mean.
03

Calculate the waiting time

The maximum guaranteed wait time can be calculated using the formula for Z-Score: \[ Z = \frac{X - µ}{σ} \] where, Z=Z-Score, X=variable (maximum guaranteed waiting time in this case), µ=mean, and σ=standard deviation. We know Z, µ and σ, so we can find X using the rearranged formula: \[ X = Z * σ + µ \]. By substituting the values \[ X = 1.645 * 2.4 + 15 \]. It gives approximately X = 18.9
04

Round up to nearest whole number

The management might want to use a whole number for simplicity. The serving time calculated is in minutes, so it makes sense to round it off to the nearest whole number. When we round up 18.9, we get 19. So, the maximum guaranteed wait time should be 19 minutes.

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Key Concepts

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

Z-Score
Imagine you want to find out how your car’s service time compares to others. That’s where the Z-score comes in handy. The Z-score indicates how many standard deviations a particular data point is from the mean.

For instance, in the given exercise, the mean service time is 15 minutes, and a standard deviation is given. If your car's service time is significantly different, the Z-score quantifies this difference.

To finding the Z-score, you'll use the formula: \( Z = \frac{X - \mu}{\sigma} \)
where \(X\) is your service time, \(\mu\) is the mean (15 minutes), and \(\sigma\) is the standard deviation (2.4 minutes).
  • Values of \( Z \) around 0 indicate times close to the mean.
  • Positive values show it took longer than most cars.
  • Negative values mean it was faster.
Understanding these scores is essential when calculating probabilities or setting guarantees like in Fast Auto Service.
Mean and Standard Deviation
Let's break down these core concepts: mean and standard deviation. The mean is simply the average of a set of numbers. In our scenario, it represents the average time for servicing a car, set at 15 minutes.

An average gives a central value, but it doesn't tell you how scattered or spread out the service times really are. That’s where standard deviation comes in.

The standard deviation measures this spread. A lower standard deviation indicates the service times are clustered closely around the mean. A higher standard deviation means more variability.

In our example, the deviation is 2.4 minutes, meaning that most service times are within 2.4 minutes of the average. Using these, Fast Auto Service can better manage wait times and set realistic expectations for customers.
Statistical Computation
Statistical computation involves crunching the numbers to make sense of data patterns. It’s the backbone of our exercise, where it helps find that maximum guaranteed waiting time for Fast Auto Service.

After understanding the problem, we had to use statistical tables or software to find a Z-score corresponding to 95% of the data. In this context, the Z-score was approximately 1.645.

Using this score, a computation translates the Z-score back into minutes so we can determine a specific wait time. The application of the formula \( X = Z \times \sigma + \mu \) allows us to compute the precise waiting time. This complex procedure joins statistical theory with practical application to solve real-world problems.
Probability
Probability helps us understand how likely events are within a certain framework. In a normal distribution, probabilities are used to determine how likely any given set of service times are. This garage wants to be sure only 5% of their customers wait longer than a certain time.

For a normal distribution:
  • About 68% of data falls within one standard deviation from the mean.
  • 95% is within two standard deviations.
  • 99.7% is within three standard deviations.
The problem specifies wanting to limit the wait time for at most 5% of the customers, meaning 95% should be serviced faster. Understanding and applying these probabilities ensures the garage sets an accurate guarantee.

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Most popular questions from this chapter

Johnson Electronics makes calculators. Consumer satisfaction is one of the top priorities of the company's management. The company guarantees the refund of money or a replacement for any calculator that malfunctions within two years from the date of purchase. It is known from past data that despite all efforts, \(5 \%\) of the calculators manufactured by this company malfunction within a 2 -year period. The company recently mailed 500 such calculators to its customers. a. Find the probability that exactly 29 of the 500 calculators will be returned for refund or replacement within a 2-year period. b. What is the probability that 27 or more of the 500 calculators will be returned for refund or replacement within a 2 -year period? c. What is the probability that 15 to 22 of the 500 calculators will be returned for refund or replacement within a 2 -year period?

Find the following probabilities for the standard normal distribution. a. \(P(z<-2.34)\) b. \(P(.67 \leq z \leq 2.59)\) c. \(P(-2.07 \leq z \leq-.93)\) d. \(P(z<1.78)\)

Find the area under the standard normal curve a. to the right of \(z=1.36\) b. to the left of \(z=-1.97\) c. to the right of \(z=-2.05\) d. to the left of \(z=1.76\)

Find the value of \(z\) so that the area under the standard normal curve a. from 0 to \(z\) is (approximately) \(.1965\) and \(z\) is positive b. between 0 and \(z\) is (approximately) 2740 and \(z\) is negative c. in the left tail is (approximately) \(.2050\) d. in the right tail is (approximately). 1053

Mong Corporation makes auto batteries. The company claims that \(80 \%\) of its LL 70 batteries are good for 70 months or longer. a. What is the probability that in a sample of 100 such batteries, exactly 85 will be good for 70 months or longer? b. Find the probability that in a sample of 100 such batteries, at most 74 will be good for 70 months or longer. c. What is the probability that in a sample of 100 such batteries, 75 to 87 will be good for 70 months or longer? d. Find the probability that in a sample of 100 such batteries, 72 to 77 will be good for 70 months or longer.

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