Refer to Exercise $79$.

(a) Find the median by hand. Show your work. Interpret your result in context.

(b) Suppose Joey has an unexcused absence for the ${15}^{th}$ quiz, and he receives a score of zero. Recalculate the mean and the median. What property of measures of center does this illustrate?

Given data:

The number of observations, n = $14$

The formula used:$M=\frac{N+1}{2}th$observation

There is no center of observation for an even number of data variables. In the data, there is a center pair, $84$and $86$, with six observations before them and six observations after them in the order list. The average of these two observations is the median.

Median

M = $\frac{84+86}{2}$

M = $85$

Or Median can be calculated as follows:

Median

M = $\frac{N+1}{2}th$observation

M = $\frac{14+1}{2}$

M = $7.5th$Observation

M = $\frac{7.5thObservation+8thobservation}{2}$

M = $\frac{84+86}{2}$

M = $85$

Interpretation: $50\%$of Joey's quiz grades are below$85$and $50\%$of Joey's quiz grades are above $85$. Therefore, the median is $85$

Mean:

The values now include $0$

$86878476919675827890809874930$

Mean = $\overline{X}$

Mean =$79.33$

Median:

Order all given data values:

$07475767880828486879091939698$

The median, or middle value of the sorted data collection, is $84$because the number of data items is odd. It's worth noting that the mean fell from $85$to $79.33$, while the median fell from $85$to $84$. The outlier"$0$" had a much greater impact (reduction) on the mean than on the median, demonstrating that the median is resistant but the mean is not. As a result, the median is $84$and the mean is $79.33$The median is resistant to change, whereas the mean is not.