Question

1. ten people donated money to a local school charity. one donated $201 and the remaining people donated $1 each. calculate the mean, median, and mode. (a) state the mean, median, and mode giving both the term and the amount. (b) which measure of central tendency should not be used to represent the general donation made to the school charity and why?

a) (a) mean = $21, median = $1, mode = $1. (b) do not use the mode — it doesn’t reflect total dollars raised.
b) (a) mean = $20, median = $1, mode = $1. (b) do not use the median — it ignores the large donation.
c) (a) mean = $21, median = $1, mode = $201. (b) do not use the mode — most gave $1 so it’s not useful.
d) (a) mean = $21, median = $1, mode = $1. (b) do not use the mean — the outlier ($201) skews it.

Explanation:

Step1: Calculate the mean

The sum of donations is \(201+(9\times1)=201 + 9=210\). There are \(n = 10\) people. The mean formula is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). So, \(\bar{x}=\frac{210}{10}=21\).

Step2: Calculate the median

Arrange the data in ascending order: \(1,1,1,1,1,1,1,1,1,201\). Since \(n = 10\) (even), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th values. The 5th and 6th values are both \(1\), so the median is \(\frac{1 + 1}{2}=1\).

Step3: Calculate the mode

The mode is the value that appears most frequently. Here, \(1\) appears \(9\) times, so the mode is \(1\).

Step4: Analyze the measure of central tendency

The mean is affected by the outlier (\(201\)). The formula for the mean \(\bar{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}\) shows that an extreme value (outlier) can skew the result. The median and mode are not affected as much by the outlier in this case.

Answer:

D. (a) Mean = \$21, Median = \$1, Mode = \$1. (b) Do not use the mean — the outlier (\$201) skews it.