waiting times (in minutes) of customers in a bank where all customers enter a single waiting line and a bank…

waiting times (in minutes) of customers in a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. find the mean and median for each of the two samples, then compare the two sets of results.\nsingle line 6.5 6.6 6.7 6.8 7.0 7.2 7.5 7.6 7.6 7.6\nindividual lines 4.3 5.3 5.8 6.2 6.6 7.6 7.6 8.6 9.3 9.8\nthe mean waiting time for customers in a single line is minutes.\n(type an integer or decimal rounded to two decimal places as needed.)

waiting times (in minutes) of customers in a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. find the mean and median for each of the two samples, then compare the two sets of results.\nsingle line 6.5 6.6 6.7 6.8 7.0 7.2 7.5 7.6 7.6 7.6\nindividual lines 4.3 5.3 5.8 6.2 6.6 7.6 7.6 8.6 9.3 9.8\nthe mean waiting time for customers in a single line is minutes.\n(type an integer or decimal rounded to two decimal places as needed.)

Answer

Explanation:

Step1: Calculate the mean for single - line data

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For the single - line data $x = {6.5,6.6,6.7,6.8,7.0,7.2,7.5,7.6,7.6,7.6}$, $n = 10$, and $\sum_{i=1}^{10}x_{i}=6.5 + 6.6+6.7+6.8+7.0+7.2+7.5+7.6+7.6+7.6=71.1$. So, $\bar{x}_{1}=\frac{71.1}{10}=7.11$.

Step2: Calculate the median for single - line data

Since $n = 10$ (an even number), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered values. The ordered data is $6.5,6.6,6.7,6.8,7.0,7.2,7.5,7.6,7.6,7.6$. The 5th value is $7.0$ and the 6th value is $7.2$. So, $Median_{1}=\frac{7.0 + 7.2}{2}=7.1$.

Step3: Calculate the mean for individual - lines data

For the individual - lines data $x={4.3,5.3,5.8,6.2,6.6,7.6,7.6,8.6,9.3,9.8}$, $n = 10$, and $\sum_{i = 1}^{10}x_{i}=4.3+5.3+5.8+6.2+6.6+7.6+7.6+8.6+9.3+9.8=71.1$. So, $\bar{x}_{2}=\frac{71.1}{10}=7.11$.

Step4: Calculate the median for individual - lines data

Since $n = 10$ (an even number), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered values. The ordered data is $4.3,5.3,5.8,6.2,6.6,7.6,7.6,8.6,9.3,9.8$. The 5th value is $6.6$ and the 6th value is $7.6$. So, $Median_{2}=\frac{6.6 + 7.6}{2}=7.1$.

Step5: Compare the results

The means of the two samples are the same ($\bar{x}{1}=\bar{x}{2}=7.11$), and the medians are also close ($Median_{1}=7.1$ and $Median_{2}=7.1$).

Answer:

The mean waiting time for customers in a single line is $7.11$ minutes. The median waiting time for customers in a single line is $7.1$ minutes. The mean waiting time for customers in individual lines is $7.11$ minutes. The median waiting time for customers in individual lines is $7.1$ minutes. The means of the two samples are equal, and the medians are close.