Q 81. - The Practice of Statistics for AP

Question

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? 

Step-by-Step Solution

Answer

Part (a) The Median = $85$

Part (b) Mean is $79.333$ and the median is $84$

Part (a) Step 1: Given information

Given data: 

86	84	91	75	78	80	74
87	76	96	82	90	98	93

The number of observations, n = $14$

Part (a) Step 2: Concept

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

Part (a) Step 3: Calculation

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 = $\frac{7.5th Observation+8th observation}{2}$

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 $85$

Part (b) Step 1: Calculation

Mean:

The values now include $0$

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

Mean = $\overline{X}$

$$\begin{gathered} =\Sigma Xi/n \\ =\frac{sum of observations}{n } \\ =\frac{x_1+x_2+x_3+....+x_n}{n} \\ =\frac{ 86+87+84+76+91+96+75+82+78+90+80+98+74+93 }{15} \\ = \frac{1190}{15} \end{gathered}$$

Mean =$79.33$

Median:

Order all given data values:

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

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.