Question

in the past month, ryan has had five pizzas delivered. he kept a record of how long it took to get the pizzas: 20 minutes, 28 minutes, 35 minutes, 40 minutes, and 25 minutes.

if the sixth time he orders pizza, it takes 95 minutes, how will this affect his data?

a.) it will make both the mean and the median less useful measures of center.

b.) it will have little to no effect on the mean, but it will make the median a less useful measure of center.

c.) it will have little to no effect on both the mean and the median.

d.) it will have little to no effect on the median, but it will make the mean a less useful measure of center.

Explanation:

Analyze the initial dataset

The initial dataset consists of five delivery times (in minutes):

$$ \{20, 28, 35, 40, 25\} $$

To find the median, we first arrange the data in ascending order:

$$ \{20, 25, 28, 35, 40\} $$

The number of observations is \(n = 5\). The median is the middle value (the 3rd value):

$$ \text{Median}_{\text{initial}} = 28 $$

The mean is calculated by summing the values and dividing by 5:

$$ \text{Mean}_{\text{initial}} = \frac{20 + 25 + 28 + 35 + 40}{5} = \frac{148}{5} = 29.6 $$

Analyze the dataset with the new observation

A sixth delivery time of 95 minutes is added. The new dataset is:

$$ \{20, 25, 28, 35, 40, 95\} $$

The value 95 is significantly larger than the rest of the data, making it an outlier.

Let's calculate the new median and mean.
With \(n = 6\) observations, the median is the average of the two middle values (the 3rd and 4th values in the ordered list):

$$ \text{Median}_{\text{new}} = \frac{28 + 35}{2} = 31.5 $$

The new mean is:

$$ \text{Mean}_{\text{new}} = \frac{20 + 25 + 28 + 35 + 40 + 95}{6} = \frac{243}{6} = 40.5 $$

Compare the effects on mean and median

The median changed slightly from 28 to 31.5 (an increase of 3.5 minutes). The median remains a reliable measure of the center of the typical delivery times.

The mean changed significantly from 29.6 to 40.5 (an increase of 10.9 minutes). Because the mean is highly sensitive to extreme values, the outlier (95 minutes) pulls the mean upward, making it much higher than almost all the actual delivery times in the dataset. Thus, the mean becomes a less useful measure of center for describing typical delivery times.

Match with the correct option

• Option (a) is incorrect because the median remains a useful measure of center.
• Option (b) is incorrect because the outlier has a large effect on the mean, not the median.
• Option (c) is incorrect because the mean is significantly affected.
• Option (d) correctly states that the outlier has little to no effect on the median, but makes the mean a less useful measure of center.

Answer:

• a.) It will make both the mean and the median less useful measures of center.
• b.) It will have little to no effect on the mean, but it will make the median a less useful measure of center.
• c.) It will have little to no effect on both the mean and the median.
• d.) It will have little to no effect on the median, but it will make the mean a less useful measure of center. (Correct answer)