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.

c.

To determine

To make: A box and whisker plot of the data.

Answer to Problem 8Q

Interquartile Range $=29.5$ , skewed right.

Explanation of Solution

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

Calculation:

  

Interquartile Range $=127.5-98=29.5$

Interquartile range implies that general spread of the data, in this case, the rough spread in the prices of bicycles is $29.5$ (difference in median of upper quartile and lower quartile).

As the median is inclined towards left side of the box, most of the data lies on left side, distribution is positive or skewed right.

d.

To determine

To find: The quantities as in part (a) when the store offers a $5\%$ discount.

Answer to Problem 8Q

  $Mean=115.9,Median=104.025,Mode=93.1,Range=110.2,S\tan dardDeviation=34.2$.

Explanation of Solution

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

Calculation:

As the discount offered is $5\%$ , all the values in the dataset become $95\%$ of their initial value, this means that all values of central tendencies also have to multiplied by $0.95$ to derive the new values,

Mean $=0.95\times 122=115.9$

Median $=0.95\times 109.5=104.025$

Mode $=0.95\times 98=93.1$

Standard deviation $(\sigma )=0.95\times 36=34.2$

Range $=0.95\times 116=110.2$