1. Calculate and interpret measures of center.
   • Learn the notation for a mean and for the sample size.
   • Calculate mean, median, and mode.
   • Understand what is meant by a “measure of center”.
   • Determine which measure of center to use for a given dataset.

Idea: What values are most common or most likely?

Three ways to measure:

• mode
• mean
• median

Mode: the most commonly occurring value.

• Used when working with categorical variables.
• We can get this easily from a frequency distribution.

Mean: this is what we usually think of as the “average”.

• Denoted \(\bar{x}\).
• Add up all of the values and divide by the number of observations (\(n\)): \[ \bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n} = \sum_{i=1}^n \frac{x_i}{n} \]
• \(x_i\) is the \(i\)th observation a
• \(\sum_{i=1}^n\) is the sum of all observations from 1 through \(n\).
  • This is called summation notation.

Median: the middle number when the data are ordered from smallest to largest.

• If there are an odd number of observations, this will be the number in the middle: {1, 3, 7, 9, 9} has median 7
• If there are an even number of observations, there will be two numbers in the middle. The median will be their average. {1, 2, 4, 7, 9, 9} has median \(\frac{4+7}{2}=5.5\)

The mean is sensitive to extreme values and skew. The median is not!

Changing that 9 out for a 45 changes the mean a lot! But the median is 7 for both \(x\) and \(y\).

Because the median is not affected by extreme observations or skew, we say it is a resistant measure or that it is robust.

• Mean: symmetric, numeric data
• Median: skewed, numeric data
• Mode: categorical data

Note: If the mean and median are roughly equal, it is reasonable to assume the distribution is roughly symmetric.

Sometimes we have reason to calculate a weighted mean.

\[\bar{x}_w = w_1x_1 + w_2x_2 + \dots w_nx_n\] Each observation is multiplied by a corresponding weight, \(w\).

• In general, each \(w\) represents the proportion attributed to that category.
• In general, we require that all of the \(w\) values sum to \(1\).

If all the weights are equal, we get an (unweighted) mean:

• We would need \(n\) equal weights which sum to 1.
  • So each weight would be \(1/n\)

Here’s what that looks like:

\[\frac{1}{n}x_1 + \frac{1}{n}x_2 + \dots + \frac{1}{n}x_n = \frac{x_1 + x_2 + \dots + x_n}{n} = \bar{x}\]

Consider the following grade distribution:

• Assignments: 15%
• Quizzes: 20%
• Exam 1: 15%
• Exam 2: 15%
• Project: 15%
• Final Exam: 20%

We can use this to calculate an overall grade.

Suppose some student has the following score in each category

• Assignments: 92%
• Quizzes: 76%
• Exam 1: 56%
• Exam 2: 69%
• Project: 89%
• Final Exam: 70%

We can calculate their overall grade in the class using the weighted average formula.

• The proportion of the grade that comes from each category is the weight.
  • We will need to convert the percentages to proportions.
• The grades are our \(x\) values.

\[\begin{aligned}\text{grade} &= 92(0.15) + 76(0.20) + 56(0.15) + 69(0.15) + 89(0.15) + 70(0.20) \\ &= 13.8 + 15.2 + 8.4 + 10.35 + 13.35 + 14 \\ &= 75.1\end{aligned}\]

So this student would get a 75.1% in the class.

Now suppose a student has the following scores

• Assignments: 83%
• Quizzes: 71%
• Exam 1: 61%
• Exam 2: 68%
• Project: 91%

and has not taken the final exam yet. He really wants to pass the class with at least a C-, but is not sure what kind of final exam grade would allow him to do that.

• If he wants to pass, he needs a minimum overall grade of 70%.
  • (He needs his weighted average to be 70% or higher.)
• We know everything except his final exam score, so we’ll make that \(F\) in our formula:

\[70 = 83(0.15) + 71(0.20) + 61(0.15) + 68(0.15) + 91(0.15) + F(0.20)\]

To figure out what he needs to get on the final, we need to solve for F.

\[\begin{aligned} 70 &= 83(0.15) + 71(0.20) + 61(0.15) + 68(0.15) + 91(0.15) + F(0.20)\\ 70 &= 12.45+14.2+9.15+10.2+13.65+0.2F \\ 70 &= 59.65 + 0.2F \\ 10.35 &= 0.2F \\ F &= 51.75 \end{aligned}\]

So he needs to get at least a 51.75% on the final exam in order to pass the class.