Chapter 1.3, Problem 81E

(a)

To determine

To find: the median

Answer to Problem 81E

Median is 85.

Explanation of Solution

Calculation:

For an even number of data values, there is no center of observation.

There is a center pair i.e. 84 and 86 in the data which have 6 observations before them and 6 observations after them in the order list. The median is the average of these two observations.

Median

  $\begin{array}{l}M=\frac{84+86}{2}\\ M=85\end{array}$

Or Median can be calculated as follows:

Median

  $\begin{array}{l}M=\frac{N+1}{2}\text{th observation}\\ M=\frac{14+1}{2}\\ M=7.5\text{th Observation }\\ M=\frac{7.5\text{th Observation}+8\text{th observation}}{2}\\ M=\frac{84+86}{2}\\ M=85\end{array}$

Interpretation: 50% of Joey's quiz grades are below 85 and 50% of Joey's quiz grades are above 85.

Conclusion:

“Therefore, the median is 85.”

(b)

To determine

To recalculate: the mean and the median if J receives a zero score.

Answer to Problem 81E

Mean is 79.33 and the median is 84.

Explanation of Solution

Calculation:

Mean:

The values now include 0:

  $\text{86 87 84 76 91 96 75 82 78 90 80 98 74 93 0}$

  $\begin{array}{c}\text{Mean}=\overline{X}\\ =\frac{{\displaystyle \sum X_i}}{n}\\ =\frac{\text{sum}\text{of}\text{observations}}{n}\\ =\frac{x_1+x_2+x_3+\mathrm{...}+x_n}{n}\\ =\frac{86+87+84+76+91+96+75+82+78+90+80+98+74+93}{15}\\ =\frac{1190}{15}\\ \text{Mean}=79.33\end{array}$

Median:

Order all given data values:

  $\text{0 74 75 76 78 80 82 84 86 87 90 91 93 96 98}$

Since the number of data values is odd, the median is the middle value of the sorted data set is 84

It is noted that the mean decreased from 85 to 79.33, while the median decreases from 85 to 84. The mean was a lot more affected (decreased) by the outlier”0” than the median, which illustrates that the median is resistant while the mean is not resistant.

Conclusion:

“Therefore, the mean is 79.33 and the median is 84.

The median is resistant, while the mean is not resistant.”