3.15 A bank branch located in a commercial district of a city has developed an improved process for A bank branch located in a commercial district of a city has developed an improved process for serving customers during the noon-to-1:00 p.m. lunch period. The waiting time, in minutes (defined as the time the customer enters the line to when he or she reaches the teller window), of a sample of 15 customers during this hour is recorded over a period of one week. The results are contained in the data file Bank1 and are listed below:
Waiting Time
4.21
5.55
3.02
5.13
4.77
2.34
3.54
3.2
4.5
6.1
0.38
5.12
6.46
6.19
3.79
a. Compute the mean and median.
b. Compute the variance, standard deviation, range, coefficient of variation, and Z scores. Are there any outliers? Explain.
c. Are the data skewed? If so, how?
"d. As a customer walks into the branch office during the lunch hour, she asks the branch manager how long she can expect to wait. The branch manager replies, “Almost
certainly less than five minutes.” On the basis of the results of (a) through (c), evaluate the accuracy of this statement."
I goofed. Now I see that the values are probably minutes and decimal portions of minutes. This invalidates my first statement…ONLY.
To answer the questions, let's first organize the data in ascending order:
0.38, 2.34, 3.02, 3.20, 3.54, 3.79, 4.21, 4.50, 4.77, 5.12, 5.13, 5.55, 6.10, 6.19, 6.46
a. Compute the mean and median:
Mean = (0.38 + 2.34 + 3.02 + 3.20 + 3.54 + 3.79 + 4.21 + 4.50 + 4.77 + 5.12 + 5.13 + 5.55 + 6.10 + 6.19 + 6.46) / 15 = 4.4607
Median = (3.54 + 4.21) / 2 = 3.875
b. Compute the variance, standard deviation, range, coefficient of variation, and Z scores:
Variance = ((0.38 - 4.4607)^2 + (2.34 - 4.4607)^2 + ... + (6.46 - 4.4607)^2) / 15 = 2.3062
Standard Deviation = sqrt(2.3062) ≈ 1.5172
Range = 6.46 - 0.38 = 6.08
Coefficient of Variation = (1.5172 / 4.4607) * 100 ≈ 34.03%
To determine if there are any outliers, we can calculate the Z scores:
Z score = (Data Point - Mean) / Standard Deviation.
Z score of 0.38: (0.38 - 4.4607) / 1.5172 ≈ -2.67
Z score of 2.34: (2.34 - 4.4607) / 1.5172 ≈ -1.40
...
Z score of 6.46: (6.46 - 4.4607) / 1.5172 ≈ 1.32
There are no outliers since all the Z scores are within the range of -3 to +3.
c. To determine if the data is skewed, we can look at the mean, median, and mode. If the mean is greater than the median, it indicates positively skewed data. If the mean is less than the median, it indicates negatively skewed data.
In this case, the mean (4.4607) is slightly greater than the median (3.875), indicating positively skewed data.
d. Based on the results of (a) through (c), the accuracy of the branch manager's statement that customers can almost certainly expect to wait less than five minutes is questionable. The mean waiting time is 4.4607 minutes and the median is 3.875 minutes, which suggests that some customers may have longer waiting times. Additionally, the positively skewed data implies that there could be a few customers with disproportionately long waiting times.
To answer the questions, we need to perform calculations on the given data. Here's how you can calculate the requested values:
a. Compute the mean and median:
- Mean: Add up all the waiting times and divide by the total number of customers (15).
(4.21 + 5.55 + 3.02 + 5.13 + 4.77 + 2.34 + 3.54 + 3.2 + 4.5 + 6.1 + 0.38 + 5.12 + 6.46 + 6.19 + 3.79) / 15 = 4.37
- Median: Arrange the waiting times in ascending order and find the middle value. In this case, since there are 15 values, the median is the 8th value.
Arrange: 0.38, 2.34, 3.02, 3.20, 3.54, 3.79, 4.21, 4.50, 4.77, 5.12, 5.13, 5.55, 6.10, 6.19, 6.46
Median: 4.77
b. Compute the variance, standard deviation, range, coefficient of variation, and Z scores:
- Variance: Calculate the average of the squared differences between each waiting time value and the mean.
Variance = Sum of (Waiting Time - Mean)^2 / (Number of Customers - 1)
Variance = [(4.21 - 4.37)^2 + (5.55 - 4.37)^2 + (3.02 - 4.37)^2 + ... + (3.79 - 4.37)^2] / 14
Variance = 2.0343 (rounded to four decimal places)
- Standard Deviation: Take the square root of the variance.
Standard Deviation = √(Variance) = √(2.0343) = 1.426 (rounded to three decimal places)
- Range: Find the difference between the largest and smallest waiting time values.
Range = Largest Waiting Time - Smallest Waiting Time
Range = 6.46 - 0.38 = 6.08
- Coefficient of Variation: The standard deviation divided by the mean, multiplied by 100%.
Coefficient of Variation = (Standard Deviation / Mean) * 100
Coefficient of Variation = (1.426 / 4.37) * 100 = 32.597% (rounded to three decimal places)
- Z-Scores: Z-score measures the number of standard deviations a data point is from the mean.
Z-Score = (Individual Value - Mean) / Standard Deviation
For each individual waiting time value, calculate their Z-score using the mean and standard deviation.
To determine if there are any outliers, we can use the Z-score method. Generally, a Z-score greater than +3 or less than -3 can be considered an outlier.
c. To determine if the data is skewed, we need to calculate the skewness of the data distribution. However, for a sample size of 15, it is not advisable to calculate skewness as it may not be accurate. In general, a larger sample size is required to make a reliable determination of skewness.
d. The branch manager stated that a customer could expect to wait "almost certainly less than five minutes." Based on the mean waiting time we calculated in part a (4.37 minutes) and the accuracy of that statement will largely depend on the tolerable level of waiting time for customers. If a customer is comfortable waiting for a few minutes more than 4.37, then the statement would be accurate. However, if the waiting time is expected to be strictly less than 5 minutes, the statement may not be accurate for all customers.
Remember to round the values appropriately based on decimal places required in the question.
Assuming the values are all in seconds, all the values are less than 5 minutes (300 seconds).
Median = point at which half the scores are below and half above.
The mean = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Range = highest score - lowest
Z = (score-mean)/SD
I'll let you do the calculations.