91影视

The waiting times are recorded by the customers.

The formula for the mean is:

$\stackrel{炉}{x}=\frac{鈭�x}{n}$, where xrepresents the observations and ndenotes the count of the observations.

Substitute the values for a single line.

$$\begin{gathered} {\stackrel{炉}{x}}_S=\frac{390+396+402+...+462}{10} \\ =\frac{4290}{10} \\ 鈮�429 \end{gathered}$$

Substitute the values for individual lines.

$$\begin{gathered} {\stackrel{炉}{x}}_I=\frac{252+324+348+...+600}{10} \\ =\frac{4290}{10} \\ =429 \end{gathered}$$

Thus, the mean values for single and individual lines are the same, i.e., 429.0 seconds.

To compute the median, find the middle values as per the total counts of the observations.

If the counts are even, the average of the middle values is the median.

Compute the median for the waiting lines of single lines.

The number of observations is10.

Arrange the observations in ascending order.

The middlemost observations are 426 and 438.

The median is given as:

$$\begin{gathered} M=\frac{426+438}{2} \\ =\frac{864}{2} \\ =432 \end{gathered}$$

Compute the median for individual measurements.

The number of observations is 10.

Arrange the observations in ascending order.

The middlemost observations are 402 and 462.

The median is given as:

$$\begin{gathered} M=\frac{402+462}{2} \\ =\frac{864}{2} \\ =432 \end{gathered}$$

Thus, the median values for single and individual lines are the same, 432.0 seconds.

The summarization of the two results gives:

The results are identical in both groups.

The waiting times for a single line have a minimum value of 390 and a maximum value of 462. On the other hand, the waiting times for individual lines vary between 252 and 600.4

The level of variation between the extreme values is large.

This characteristic of the data is not clear from the measure of the center. It can be established using the measures of variation.