## Related questions with answers

Check each data set for outliers.

a. $14,18,27,26,19,13,5,25$ b. $112,157,192,116,153,129,131$

Solution

Verified$\text{\textcolor{#4257b2}{\textbf{(a) 14, 18, 27, 26, 19, 13, 5, 25 }}}$

To find the outliers, we execute following steps :

$\textbf{Step 1 : }$ Computing median :

First we arrange the data-set in order from lowest to highest :

$5\text{ , }13\text{ , }14\text{ , }18\text{ , }19\text{ , }25\text{ , }26\text{ , }27$

$\text{number of values} = n = 8$

Since number of values are even hence the median is mean of $\left(\dfrac{n}{2}\right)^{th}$ and $\left(\dfrac{n+2}{2}\right)^{th}$ value. Hence median of data set is mean of $4^{th}$ and $5^{th}$ value.

$\begin{align*} \text{median} &= \left(\dfrac{18+19}{2}\right)\\ \text{median} &= 18.5 \end{align*}$

$\textbf{Step 2 : }$ Computing $Q_{1}$ :

$Q_{1}$ for a given data set is the median of values less than median of that data set.

The new data-set formed using values less than median $(=18.5)$ is :

$5\text{ , }13\text{ , }14\text{ , }18$

$\text{number of values in new data set} = n = 4$

Since number of values are even hence the median is mean of $\left(\dfrac{n}{2}\right)^{th}$ and $\left(\dfrac{n+2}{2}\right)^{th}$ value. Hence median of data set is mean of $2^{nd}$ and $3^{rd}$ value.

$\begin{align*} Q_{1} &= \left(\dfrac{13+14}{2}\right)\\ Q_{1} &= 13.5 \end{align*}$

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