91影视

We will use the z-score method to find the outliers. For that, we will first arrange the data in ascending order.

(32, 33, 39, 43, 44, 49, 49, 50, 50, 51, 51, 54, 56, 59, 63, 64, 64, 65, 68, 71, 73, 82, 86, 95, 102)

In order to calculate the z-score we require mean and the standard deviation. So next we will calculate that.

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{observations}}{\mathrm{No}.\mathrm{of}\mathrm{observations}}\\ & =& \frac{1493}{25}\\ & =& 59.72\end{array}$

$$\begin{gathered} Variance=\frac{鈭�\left(x-\stackrel{炉}{x}\right)}{n-1} \\ =\frac{{\left(\text{3}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{3}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{39}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{43}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{44}-\text{59.72}\right)}^{\text{2}} \\ \text{+}{\left(\text{49}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{49}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{50}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{50}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{51}-5\text{9.7}2\right)}^{\text{2}} \\ \text{+ (51}-5{\text{9.72)}}^{\text{2}}\text{+}{\left(\text{54}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{56}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{59}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{63}-\text{59.72}\right)}^{\text{2}} \\ \text{+}{\left(\text{64}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{64}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{65}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{68}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{71}-\text{59.72}\right)}^{\text{2}} \\ \text{+}{\left(\text{73}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{73}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{82}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{86}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{95}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{102}-\text{59.72}\right)}^{\text{2}}}{\text{25 - 1}} \\ \text{=}\frac{7723.04}{24} \\ =321.79 \end{gathered}$$

$\begin{array}{rcl}\mathrm{Standard}\mathrm{Deviation}& =& \sqrt{\mathrm{Variance}}\\ & =& \sqrt{321.79}\\ & =& 17.94\end{array}$

Scores whose z score lies beyond -3 and 3 are outliers

$\begin{array}{rcl}z& =& \frac{x-渭}{未}\\ & =& \frac{32-59.72}{17.94}\\ & =& -1.54\end{array}$

Since the z score for 32 is -1.54 it is not an outlier.

$\begin{array}{rcl}z& =& \frac{x-渭}{未}\\ & =& \frac{95-59.72}{17.94}\\ & =& 1.966\end{array}$

As the z score of 95 is 1.966 it is not an outlier.

$\begin{array}{rcl}z& =& \frac{x-渭}{未}\\ & =& \frac{102-59.72}{17.94}\\ & =& 2.35\end{array}$

Since the z score of 102 is 2.35 it is not an outlier but a suspect outlier.

Therefore, there are no outliers in this data set.

It is observed that the z scores of repeat customer鈥檚 lies well within 2 standard deviations from mean and the scores of onetime customers is close to or beyond 2 standard deviations from mean. Therefore, we can say that the delivery time for repeat customers is indeed less.