Measures of Central Tendency Mode, Median, Mean

The Mode The mode of a data set is the value that occurs most frequently. Example (3.1: Exercise 2). Number of home runs Babe Ruth hit in his 12 Home run titles The mode for this data is 46.

The Median The median is the central value in an ordered distribution. To find it 1.Order the data from smallest to largest. 2.For an odd number of data values in a distribution, Median = Middle data value 3.For an even number of data values in a distribution, Median = (sum of middle two values)/2

Example. (a) Find the median number of home runs Babe Ruth hit in his 12 home run titles (b) Find the median if the highest number 60 is removed from the list.

Solution. (a) Order the data There are an even number of data, the average of the middle two data is 46.5 (b) If 60 is removed from the list, then the median is in the 6th place which is 46.

In general, in a list of ordered data with n numbers, the median is in the (n+1)/2 position. For example, in an ordered list with 75 numbers, the median would be in the (75+1)/2 = 38th position. In an ordered list of 20 numbers, the median would be in the 10.5th position, which means to average the 10th and 11th data.

For a population with N data, the mean is For a sample with n data, the mean is In general, the mean is what we typically think of as the average.

Example (3.1 Exercise 6) Ages of a random sample of 40 pro football players. 21,21,22,22,22,22,23,23,23,23 24,24,24,25,25,25,25,25,25,25 26,26,26,26,27,27,28,28,28,29 29,29,29,29,30,31,31,32,33,37 Compute the mean, median and mode of this data. Note the sum of the data is 1050.

Solution: The mean is 1050/40 = The median is in the 20.5th position, and so the median is The mode is the most common data which is 25 (occurs 7 times in list).

A student receives grades of A,A,B,C,C and is surprised to receive a GPA of 3.5 because the average grade of A,A,B,C,C is a B since ( )/5 = 3.0 Explain how the student’s GPA could be 3.5.

Solution: GPA’s are based on a weighted average, i.e. Weighted Average = Where x is the data value and w is the weight assigned to the data value. In this case suppose the A’s were in 5 unit classes, the B in 4 unit class, and the C’s were in 1 unit classes. Then the weighted average is 4(5)+4(5) + 3(4) + 2(1) + 2(1) Which is 56/16 = 3.5

Example. Bob from “Account Temps” is pleased when his boss tells him that his salary is in the top 10% of all company salaries. Then a month later Bob learns that his salary is less than half the mean salary in the company, and Bob is very angry with his boss. Explain how Bob’s boss could have been telling the truth.

Solution. The salaries in Bob’s company with 20 employees could be: 20,20,25,25,30,30,30,30,30,40 40,40,40,40,45,45,45,50,60,2000 where Bob’s salary is 60K per year. The average annual salary is ∑x/20 = 2680/20 = 134K which is more than double Bob’s salary.

Example. The salaries in Bob’s company with 20 employees are: 20,20,25,25,30,30,30,30,30,40 40,40,40,40,45,45,45,50,60,2000 in thousand’s per year. Compute the median salary and the 5% trimmed mean.

Solution. The median salary is 40K per year. The 5% trimmed mean is 660,000/18 = 36,667 (this is obtained by removing highest 5% and lowest 5% of the data) Think about it: how much would the median change by changing the highest salary to 75K per year? How much would the mean change?

Think about it: An army crosses a river whose average depth is 1 foot, yet several soldiers drowned when they crossed it because they couldn’t swim. Explain how this could happen. Hint: consider the possibility that it is a wide river that is very shallow for most of the width, but has a deep narrow trench in which the soldiers drowned.

Think about it: Ken had an average of 50% on exams and 90% on assignments in his class, so he computed that his overall average was 70% which according to the syllabus is a C. He was shocked to see that his grade was a D. Why should Ken not have been surprised with his grade?

One Solution: If exams count for 90% of the entire grade and homework counts for 10% of the grade, then Ken’s (weighted) overall average in the class is (.90)(50%) + (.1)(90%) = 45% + 9% = 54% which is why he received a D.