Question

The number of bagels sold daily for two bakeries is shown in the table:Bakery ABakery B1515521651343335571212945364617Based on these data, is it better to describe the centers of distribution in terms of the mean or the median? Explain. Mean for both bakeries because the data is symmetric Mean for Bakery B because the data is symmetric; median for Bakery A because the data is not symmetric Mean for Bakery A because the data is symmetric; median for Bakery B because the data is not symmetric Median for both bakeries because the data is not symmetric

Answer 1

We can see from the question two data sets, one for Bakery A, and one for Bakery B. Each of them has 8 values.

We need to determine to choose which is better to describe the center of distribution in terms of the mean or the median.

To find it, we can construct a boxplot for each data set, and determine if both data sets are skewed or not. If they are skewed, the data is not symmetric, and we cannot use the mean as a central tendency measure but we can use the median instead.

Then we can proceed as follows:

1. Order data in ascending order for each data set:

Bakery A:

`\begin{gathered} \text{ Bakery A=\textbraceleft15,52,51,33,57,12,45,46\textbraceright} \\ \\ \text{ After ordering in ascending order:} \\ \\ \text{ Bakery A=\textbraceleft12,15,33,45,46,51,52,57\textbraceright} \end{gathered}`

Bakery B:

`\begin{gathered} \text{ Bakery B=\textbraceleft15, 16, 34, 35, 12,9, 36,17\textbraceright} \\ \\ \text{ After ordering in ascending order:} \\ \\ \text{ Bakery B=\textbraceleft9,12,15,16,17,34,35,36\textbraceright} \end{gathered}`

2. We need to find the mean, minimum value, the first quartile, the median, the third quartile, and the maximum value for each data set.

For Bakery A:

• Mean:

`\begin{gathered} \text{ Mean=}((12+15+33+45+46+51+52+57))/(8)=(311)/(8)=38.875 \\ \\ \text{ Mean=}38.875 \end{gathered}`

• Minimum: It is the minimum of the data set Bakery A: 12.

• Median: it is the value for which 50% of the data is below it and the other 50% is above it:

Then the median is 45.5.

• The First Quartile (Q1) and the Third Quartile (Q3) are, roughly speaking, the median of the first half of the data, and the median of the second half of the data. Then we have:

Then the First Quartile (Q1) = 24, and the Third Quartile (Q3) = 51.5.

Maximum: It is the maximum value of the data = 57.

In summary:

• Minimum: 12

,

• Q1: 24

,

• Median: 45.5

,

• Q3: 51.3

,

• Maximum: 57

Now we can draw the boxplot for Bakery A as follows:

For Bakery B

We can proceed as before, and we can find that the values for constructing a boxplot of the data are (we can use a graphing calculator to achieve that):

• Mean: 21.75

,

• Minimum: 9

,

• Q1: 13.5

,

• Median: 16.5

,

• Q3: 34.5

,

• Maximum: 36

And the corresponding boxplot is:

3. Now, we can analyze the situation as follows:

Bakery A: We can see that the distance from the median to the minimum value is greater than the distance from the median to the maximum, then we can conclude that the boxplot is negatively skewed. Then the mean is pulled to the minimum values, and because of this, the mean is NOT a good measure of central tendency - in this case. Additionally, we can see that the mean and the median are not the same, and we do not have a symmetrical distribution.

In cases when we have skewed distributions, the best measure of central tendency is the Median.

Bakery B: We can see that the distance from the median to the maximum value is greater than the distance from the median to the minimum, then we can conclude that the boxplot is positively skewed. We can see that the mean is pulled to the maximum values, and because of this, the mean is NOT a good measure of central tendency - in this case. Additionally, we can see that, also, the mean and the median are not the same, and we do not have a symmetrical distribution either.

In cases when we have skewed distributions, the best measure of central tendency is the Median.

Summary

Therefore, we can conclude that in both cases the mean is NOT a good measure of central tendency because the data is not symmetric in both cases. The mean is pulled in both cases because the distributions are skewed.

Then as a conclusion, we can say that the centers of distribution can be described in terms of median because the data is not symmetric, and we can choose the following option:

Median for both bakeries because the data is not symmetric (Answer) (last option)