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The management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes a. Find the probability that a randomly selected customer will have to wait for less than 3 minutes. b. What percentage of the customers have to wait for 10 to 13 minutes? c. What percentage of the customers have to wait for 6 to 12 minutes? d. Is it possible that a customer may have to wait longer than 16 minutes for service? Explain.

Short Answer

Expert verified
a. The probability that a customer will wait for less than 3 minutes is 0.0062 or 0.62%. b. The percentage of customers who have to wait 10 to 13 minutes is approximately 15.25%. c. The percentage of customers who have to wait 6 to 12 minutes is approximately 81.85%. d. It is possible, in theory, for a customer to wait longer than 16 minutes, even though the probability is extremely low.

Step by step solution

01

Compute the Z-score for 3 minutes

To compute Z-score use the formula \[Z = (X - \mu) / \sigma\] Here, X=3, \(\mu\)=8, and \(\sigma\)=2. So, Z-score will be \[Z = (3 - 8) / 2 = -2.5\] This Z-score indicates how many standard deviations an element is from the mean.
02

Find the probability for waiting less than 3 minutes

Look up the Z-score in the Z-table, which gives the area to the left of that Z-score. The Z-value of -2.5 is 0.0062, this represents the probability that a selected customer will have to wait for less than 3 minutes.
03

Compute Z-scores for 10 and 13 minutes

Same as step 1, compute the Z-scores for 10 and 13 minutes. The Z-scores are \(Z_{10} = (10 - 8) / 2 = 1\) and \(Z_{13} = (13 - 8) / 2 = 2.5\)
04

Find the percentage for waiting between 10 to 13 minutes

The percentage of customers having to wait between 10 to 13 minutes can be found by subtracting the Z-table values corresponding to the calculated Z-scores. From the Z-table, the values are 0.8413 for Z=1 and 0.9938 for Z=2.5. The difference \(0.9938 - 0.8413 = 0.1525\) or 15.25% is the percentage of customers having to wait between 10 to 13 minutes.
05

Compute Z-scores for 6 and 12 minutes

Compute the Z-scores for 6 and 12 minutes. The Z-scores will be \(Z_{6} = (6 - 8) / 2 = -1\) and \(Z_{12} = (12 - 8) / 2 = 2\)
06

Find the percentage for waiting between 6 to 12 minutes

From the Z-table, the values are 0.1587 for Z=-1 and 0.9772 for Z=2. The difference \(0.9772 - 0.1587 = 0.8185\) or 81.85% is the percentage of customers having to wait between 6 to 12 minutes.
07

Analysis

The bank manager might express concern if a customer has to wait longer than 16 minutes. Given a normal distribution, all events are technically possible, though they can have miniscule probabilities. By calculating the Z-score for 16 minutes, \(Z_{16} = (16 - 8) / 2 = 4\), and referring to a Z-table, the corresponding value is approximately 1, which confirms that it is possible for a customer to wait longer than 16 minutes, even though the likelihood is statistically insignificant.

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Key Concepts

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

z-score
The concept of a z-score is key in understanding normal distribution. A z-score helps us determine how far away a particular value is from the mean, measured in standard deviations. This is extremely useful to compare different data points within a normal distribution. To calculate the z-score, use the formula: \[ Z = \frac{X - \mu}{\sigma} \] Where:
  • \(X\) is the value you are evaluating.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
For example, if you want to find out how different a customer's waiting time of 3 minutes is from the average (8 minutes in this exercise), you substitute into the formula: \[ Z = \frac{3 - 8}{2} = -2.5 \] A z-score of -2.5 indicates that 3 minutes is 2.5 standard deviations below the mean.
probability
Understanding probability in the context of normal distribution means predicting the likelihood of a certain outcome. The normal distribution, which often appears as a bell-shaped curve, enables us to use z-scores to find probabilities for different outcomes. Once you have a z-score, you can use a z-table (or standard normal distribution table) to find the probability. The table tells you the probability of a value being less than the given z-score. For example, a z-score of -2.5 corresponds to a probability of 0.0062, meaning there's a 0.62% chance that a randomly selected customer will wait less than 3 minutes. Similarly, to find the probability of waiting times falling within a range, we can look up each z-score in the table and subtract the smaller probability from the larger one. For instance, the probability of waiting 10 to 13 minutes is found by subtracting the z-score probabilities of 10 minutes (0.8413) from 13 minutes (0.9938), resulting in 15.25%.
standard deviation
Standard deviation is a critical component of understanding the spread in a set of data. In the setting of a normal distribution, it denotes how much variation or "dipersion" exists from the mean. A smaller standard deviation means the data points tend to be close to the mean, while a larger standard deviation indicates more spread out data points. In the exercise scenario, the standard deviation of customer waiting times is 2 minutes. This gives us insight into how varied or consistent these times are. When calculating z-scores, the standard deviation is used to scale differences from the mean, making them unitless, which allows comparing them within the same data set. Thus, it aids in determining the likelihood of various waiting periods and how they diverge or converge around the average of 8 minutes. Understanding this spread helps the bank management anticipate and provide efficient customer service by minimizing extreme waiting times.
mean
The mean is commonly known as the average, and it's a central value around which a normal distribution is organized. It's calculated by summing up all values and dividing by the number of values. In the exercise, the mean waiting time is 8 minutes, representing the central tendency of customer waiting times at the bank. This mean serves as a reference point for calculating z-scores, which explain how each waiting time compares to the average. It's also essential in understanding the symmetrical property of normal distribution, where about 68% of values lie within one standard deviation, 95% within two, and 99.7% within three standard deviations of the mean. A normal distribution with a well-understood mean helps predict the probability of different waiting times, guiding the branch manager's efforts in improving service efficiency. Understanding the mean and its role in normal distributions enables proactive adjustments to improve customer satisfaction in terms of waiting time.

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