Chapter 11.3, Problem 8Q

a.

To determine

To find: The mean, median, mode, range and standard deviation of the prices.

Answer to Problem 8Q

  $Mean=122,Median=109.5,Mode=98,Range=116,S\tan dardDeviation=36$.

Explanation of Solution

Given Information: A table showing the prices of eight mountain bikes in a sporting goods store is given.

Calculation:

  $\begin{array}{l}Mean=\frac{98+119+95+211+130+98+100+125}{8}\\ =122\end{array}$

Arranging in ascending order,

  $95,98,98,100,119,125,130,211$

  $\begin{array}{l}Median=\frac{100+119}{2}\\ =109.5\end{array}$

The term $98$ occurred two times whereas all other terms occurred one time,

  $\Rightarrow Mode=98$

Highest number $=211$

Lowest number $=95$

  $\begin{array}{l}\Rightarrow Range=211-95\\ =116\end{array}$

  $\begin{array}{l}\sigma =\sqrt{\frac{{(95-122)}^2+2\times {(98-122)}^2+{(100-122)}^2+{(119-122)}^2+{(125-122)}^2+{(130-122)}^2+{(211-122)}^2}{8}}\\ \sigma =\sqrt{1296}\\ \sigma =36\\ \end{array}$

b.

To determine

To identify: The outlier.

Answer to Problem 8Q

Outlier is $211$ .

Explanation of Solution

Given Information: A table showing the prices of eight mountain bikes in a sporting goods store is given.

Outlier is $211$ as it is far greater than the central tendency, outlier mainly affects mean by imparting increase in its value, it doesn’t have an apparent affect on median and mode.