Effects of an Outlier Listed below are platelet counts (1000 cells>mL) from subjects included in Data Set 1 “Body Data.” Identify the outlier and then comment on the effect it has on the mean and standard deviation by finding the values of those statistics with the outlier included and then with the outlier excluded. 263 206 185 246 188 191 308 262 198 253 646

The platelet count of different subjects is in terms of $\left(1000\text{cells}/\mu L\right)$ .

263 206 185 246 188 191 308 262 198 253 646

For n observations denoted by x, the mean value is computed using the following formula.

$\overline{x}=\frac{\sum x}{n}$

The sample standard deviation is computed using the following formula.

$s=\sqrt{\frac{\sum {\left(x-\overline{x}\right)}^2}{n-1}}$

Substitute the value for mean as follows.

$$\begin{gathered} \overline{x}=\frac{263+206+185+...+646}{11} \\ =\frac{2946}{11} \\ =267.8182 \end{gathered}$$

Thus, the sample mean value is 267.8$\left(1000\text{cells}/\mu L\right)$.

Substitute the values for sample standard deviation as follows.

$$\begin{gathered} s=\sqrt{\frac{{\left(263-267.8\right)}^2+{\left(206-267.8\right)}^2+...+{\left(646-267.8\right)}^2}{11-1}} \\ =\sqrt{\frac{23.21+3821.49+...+143021.5}{10}} \\ =\sqrt{\frac{173215.64}{10}} \\ =131.6114 \end{gathered}$$

Thus, the sample standard deviation is 131.6 $\left(1000\text{cells}/\mu L\right)$.

Outliers are one or more observations that are unusual from the set of observations. It can be termed as extreme among other numerical values.

In the given set of observations, 646 is the outlier as it is extreme compared to the range of the dataset from 185 to 308.

Substitute the values for mean excluding the outlier 646 as follows.

$$\begin{gathered} \overline{x}=\frac{263+206+185+...+253}{11} \\ =\frac{2300}{10} \\ =230 \end{gathered}$$

Thus, the sample mean value is 230.0$\left(1000\text{cells}/\mu L\right)$.

Substitute the values for sample standard deviation as follows.

$$\begin{gathered} s=\sqrt{\frac{{\left(263-230\right)}^2+{\left(206-230\right)}^2+...+{\left(253-230\right)}^2}{10-1}} \\ =\sqrt{\frac{1089+576+...+529}{9}} \\ =\sqrt{\frac{15892}{9}} \\ =42.02 \end{gathered}$$

Thus, the sample standard deviation is 42.0 $\left(1000\text{cells}/\mu L\right)$.

The summary result is shown below.

Thus, both measures have changed significantly.

Specifically, the sample standard deviation has reduced by a vast measure.