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Many retail stores offer their own credit cards. At the time of the credit application the customer is given a 10 percent discount on the purchase. The time required for the credit application process follows a uniform distribution with the times ranging from 4 minutes to 10 minutes. a. What is the mean time for the application process? b. What is the standard deviation of the process time? c. What is the likelihood a particular application will take less than 6 minutes? d. What is the likelihood an application will take more than 5 minutes?

Short Answer

Expert verified
a. 7 minutes b. About 1.732 minutes c. 0.33 d. 0.83

Step by step solution

01

Understanding the Uniform Distribution

A uniform distribution means that every outcome in a given range is equally likely. In this problem, the application time is uniformly distributed between 4 minutes and 10 minutes. The range of the distribution is from 4 to 10 minutes.
02

Calculating the Mean

The mean of a uniform distribution is calculated using the formula \( \mu = \frac{a+b}{2} \), where \(a\) and \(b\) are the minimum and maximum values, respectively. Here, \(a = 4\) and \(b = 10\). \[ \mu = \frac{4+10}{2} = \frac{14}{2} = 7 \] So, the mean time for the application process is 7 minutes.
03

Calculating the Standard Deviation

The standard deviation for a uniform distribution is calculated using the formula \( \sigma = \frac{b-a}{\sqrt{12}} \). Substituting \(a = 4\) and \(b = 10\), we get: \[ \sigma = \frac{10-4}{\sqrt{12}} = \frac{6}{\sqrt{12}} = \frac{6}{2\sqrt{3}} = \sqrt{3} \approx 1.732 \] The standard deviation of the process time is approximately 1.732 minutes.
04

Calculating the Probability for Less Than 6 Minutes

For a uniform distribution, the probability that an event takes less than a specific time \(c\) is given by the area of the interval from \(a\) to \(c\) over the total interval \(b-a\). \[ P(X < 6) = \frac{6 - 4}{10 - 4} = \frac{2}{6} = \frac{1}{3} \approx 0.3333 \] Thus, the probability that the application takes less than 6 minutes is approximately 0.33.
05

Calculating the Probability for More Than 5 Minutes

Similarly, the probability that it takes more than 5 minutes is calculated by: \[ P(X > 5) = \frac{10 - 5}{10 - 4} = \frac{5}{6} \approx 0.8333 \] So, the likelihood that an application takes more than 5 minutes is approximately 0.83.

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Key Concepts

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

Mean Calculation
The mean of a uniform distribution is one of the simplest concepts to grasp in statistics. By definition, a uniform distribution implies that every outcome within a given range is equally likely. This simplicity allows us to find the mean, or average, using a straightforward formula.

To calculate the mean, you take the average of the smallest (\(a\)) and largest (\(b\)) values within the range. The formula is:
  • \( \mu = \frac{a+b}{2} \)
In this case, the minimum is 4 minutes and the maximum is 10 minutes. Plugging these values into our formula, we calculate:
  • \( \mu = \frac{4+10}{2} = 7 \)
So, for a customer applying for a credit card at this store, the average time they should expect to spend is 7 minutes.

This calculated mean gives both the store and the customer an idea of what the typical application process time may look like.
Standard Deviation
Understanding standard deviation is crucial because it provides insight into the variability of the data. It tells us how spread out the times are in the distribution. For a uniform distribution, the standard deviation can be found by using the formula:
  • \( \sigma = \frac{b-a}{\sqrt{12}} \)
Here, \(a\) is 4 minutes, and \(b\) is 10 minutes, as before. The calculations are:
  • \( \sigma = \frac{10-4}{\sqrt{12}} = \frac{6}{\sqrt{12}} = \frac{6}{2\sqrt{3}} = \sqrt{3} \)
This evaluates approximately to 1.732 minutes.

The standard deviation gives us an indication of the range within which most application times will occur. A lower standard deviation would mean less variation and more predictability in process times, whereas a higher value indicates greater variation.
Probability
Probability is about understanding the likelihood of an event occurring. In a uniform distribution, like our credit card application process, calculating probabilities is quite intuitive. For example, if you want to know the probability that an application takes less than 6 minutes, use the formula for probability in uniform distributions:
  • \( P(X < c) = \frac{c-a}{b-a} \)
Plugging in the values for less than 6 minutes, we find:
  • \( P(X < 6) = \frac{6-4}{10-4} = \frac{1}{3} \)
This means there's about a 33% chance that the application process will be completed in under 6 minutes.

Similarly, for an application taking more than 5 minutes, the probability would be:
  • \( P(X > 5) = \frac{10-5}{10-4} = \frac{5}{6} \)
This translates to approximately an 83% likelihood of taking more than 5 minutes. Understanding these probabilities helps both the business to optimize their processes and the customers to plan their time accordingly.
Statistics for Business
Statistics can be a powerful tool for businesses to analyze and improve their operations. Understanding concepts like uniform distribution, mean, and standard deviation is crucial for efficient business management. In our store's credit card application scenario, these statistical measures provide valuable insights.

For instance, the mean time can inform staffing needs during peak hours by predicting average processing times. If most applications average around 7 minutes, a manager can estimate machine or personnel needs accordingly.

Moreover, knowledge of standard deviation helps businesses measure variability in processing times. Understanding that most times fall within a range of 1.732 minutes around the mean can help with planning for unexpected surges in customer volume or technical issues.

Finally, employing probability calculations allows businesses to prepare for common scenarios. By knowing that the likelihood of processing an application in under 6 minutes is only around 33%, managers can set realistic customer service expectations and offer accurate wait time estimates.

Overall, these statistical tools enable more informed decision-making, leading to better customer service and more efficient operations.

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Most popular questions from this chapter

Assume that the mean hourly cost to operate a commercial airplane follows the normal distribution with a mean \(\$ 2,100\) per hour and a standard deviation of \(\$ 250 .\) What is the operating cost for the lowest 3 percent of the airplanes?

The amounts of money requested on home loan applications at Down River Federal Savings follow the normal distribution, with a mean of \(\$ 70,000\) and a standard deviation of \(\$ 20,000 .\) A loan application is received this morning. What is the probability: a. The amount requested is \(\$ 80,000\) or more? b. The amount requested is between \(\$ 65,000\) and \(\$ 80,000 ?\) c. The amount requested is \(\$ 65,000\) or more?

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Management at Gordon Electronics is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 percent of production based on past experience. Past records indicate weekly production follows the normal distribution. The mean of this distribution is 4,000 units per week and the standard deviation is 60 units per week. If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?

According to the Insurance Institute of America, a family of four spends between \(\$ 400\) and \(\$ 3,800\) per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts. a. What is the mean amount spent on insurance? b. What is the standard deviation of the amount spent? c. If we select a family at random, what is the probability they spend less than \(\$ 2,000\) per year on insurance per year? d. What is the probability a family spends more than \(\$ 3,000\) per year?

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