Answer :
We are given two waiting‐time configurations. One configuration corresponds to the waiting times when there are two lines, and the other corresponds to the waiting times when there is a single line. Each configuration is arranged in a 5×5 table, so there are 25 waiting times for each setup.
Below is a detailed explanation of how to compute the mean and the standard deviation for each configuration.
–––––––––––––––––––––––––––––––––––––––––––––––
Step 1. Write the Data in Matrix Form
For the two‐line configuration, the waiting times (in seconds) are given by
[tex]$$
\begin{bmatrix}
63.6 & 216.1 & 86.2 & 340.1 & 200.1 \\
629.8 & 333.0 & 328.7 & 914.8 & 553.2 \\
596.7 & 864.8 & 1090.2 & 663.2 & 517.8 \\
565.7 & 268.0 & 350.2 & 94.9 & 99.7 \\
163.2 & 101.2 & 206.1 & 504.0 & 457.4
\end{bmatrix}.
$$[/tex]
For the single‐line configuration, the waiting times (in seconds) are
[tex]$$
\begin{bmatrix}
64.3 & 157.4 & 141.7 & 279.3 & 252.8 \\
476.3 & 477.9 & 473.9 & 402.2 & 722.3 \\
761.2 & 692.3 & 837.1 & 903.4 & 734.1 \\
605.8 & 267.8 & 309.8 & 128.8 & 133.0 \\
121.7 & 129.3 & 232.9 & 460.7 & 482.3
\end{bmatrix}.
$$[/tex]
–––––––––––––––––––––––––––––––––––––––––––––––
Step 2. Compute the Mean
Recall that the mean (or average) of a set of values is given by
[tex]$$
\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}.
$$[/tex]
Since each matrix has 25 values, we add all the numbers in each configuration and then divide the resulting sum by 25.
For the two‐line configuration, the computed mean is approximately
[tex]$$
\text{Two Line Mean} \approx 408.348 \text{ seconds}.
$$[/tex]
For the single‐line configuration, the computed mean is approximately
[tex]$$
\text{Single Line Mean} \approx 409.932 \text{ seconds}.
$$[/tex]
–––––––––––––––––––––––––––––––––––––––––––––––
Step 3. Compute the Standard Deviation
The standard deviation measures how spread out the numbers are from the mean. It is computed using the formula
[tex]$$
\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2},
$$[/tex]
where [tex]$N$[/tex] is the number of values, [tex]$x_i$[/tex] represents each waiting time, and [tex]$\mu$[/tex] is the mean.
For the two‐line configuration, after calculating the squared differences from the mean for each waiting time and then taking the square root of the average of these squared differences, we obtain a standard deviation of approximately
[tex]$$
\sigma_{\text{Two Line}} \approx 274.232 \text{ seconds}.
$$[/tex]
For the single‐line configuration, the computed standard deviation is approximately
[tex]$$
\sigma_{\text{Single Line}} \approx 250.3743 \text{ seconds}.
$$[/tex]
–––––––––––––––––––––––––––––––––––––––––––––––
Step 4. Summary of the Results
The final computed statistics for the waiting times are:
1. For the two‐line configuration:
- Mean: [tex]$$408.348$$[/tex] seconds
- Standard Deviation: [tex]$$274.232$$[/tex] seconds
2. For the single‐line configuration:
- Mean: [tex]$$409.932$$[/tex] seconds
- Standard Deviation: [tex]$$250.3743$$[/tex] seconds
These results provide a summary of the average waiting times and the variability within each waiting line configuration.
Below is a detailed explanation of how to compute the mean and the standard deviation for each configuration.
–––––––––––––––––––––––––––––––––––––––––––––––
Step 1. Write the Data in Matrix Form
For the two‐line configuration, the waiting times (in seconds) are given by
[tex]$$
\begin{bmatrix}
63.6 & 216.1 & 86.2 & 340.1 & 200.1 \\
629.8 & 333.0 & 328.7 & 914.8 & 553.2 \\
596.7 & 864.8 & 1090.2 & 663.2 & 517.8 \\
565.7 & 268.0 & 350.2 & 94.9 & 99.7 \\
163.2 & 101.2 & 206.1 & 504.0 & 457.4
\end{bmatrix}.
$$[/tex]
For the single‐line configuration, the waiting times (in seconds) are
[tex]$$
\begin{bmatrix}
64.3 & 157.4 & 141.7 & 279.3 & 252.8 \\
476.3 & 477.9 & 473.9 & 402.2 & 722.3 \\
761.2 & 692.3 & 837.1 & 903.4 & 734.1 \\
605.8 & 267.8 & 309.8 & 128.8 & 133.0 \\
121.7 & 129.3 & 232.9 & 460.7 & 482.3
\end{bmatrix}.
$$[/tex]
–––––––––––––––––––––––––––––––––––––––––––––––
Step 2. Compute the Mean
Recall that the mean (or average) of a set of values is given by
[tex]$$
\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}.
$$[/tex]
Since each matrix has 25 values, we add all the numbers in each configuration and then divide the resulting sum by 25.
For the two‐line configuration, the computed mean is approximately
[tex]$$
\text{Two Line Mean} \approx 408.348 \text{ seconds}.
$$[/tex]
For the single‐line configuration, the computed mean is approximately
[tex]$$
\text{Single Line Mean} \approx 409.932 \text{ seconds}.
$$[/tex]
–––––––––––––––––––––––––––––––––––––––––––––––
Step 3. Compute the Standard Deviation
The standard deviation measures how spread out the numbers are from the mean. It is computed using the formula
[tex]$$
\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2},
$$[/tex]
where [tex]$N$[/tex] is the number of values, [tex]$x_i$[/tex] represents each waiting time, and [tex]$\mu$[/tex] is the mean.
For the two‐line configuration, after calculating the squared differences from the mean for each waiting time and then taking the square root of the average of these squared differences, we obtain a standard deviation of approximately
[tex]$$
\sigma_{\text{Two Line}} \approx 274.232 \text{ seconds}.
$$[/tex]
For the single‐line configuration, the computed standard deviation is approximately
[tex]$$
\sigma_{\text{Single Line}} \approx 250.3743 \text{ seconds}.
$$[/tex]
–––––––––––––––––––––––––––––––––––––––––––––––
Step 4. Summary of the Results
The final computed statistics for the waiting times are:
1. For the two‐line configuration:
- Mean: [tex]$$408.348$$[/tex] seconds
- Standard Deviation: [tex]$$274.232$$[/tex] seconds
2. For the single‐line configuration:
- Mean: [tex]$$409.932$$[/tex] seconds
- Standard Deviation: [tex]$$250.3743$$[/tex] seconds
These results provide a summary of the average waiting times and the variability within each waiting line configuration.
