STATISTICS

Statistics are used to describe data. Types of statistics you are likely to encounter are:

Average (arithmetic mean): The average of a set of numbers is the sum of the numbers divided by the number of numbers in the set.

Median: The median is the middle number of a set of numbers that is arranged in numerical order. (Think of the "median" of a street).

Note that if there are an even number of data points the median is the average of the two middle data points.

Mode: The mode is the most common number in a set of numbers. There can be more than one mode.

Example: Given the following data points, find the mean, the median, and the mode.

1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 4, 7.

Mean: (1+1+2+2+2+2+3+4+4+4+4+7) ¸ 12 = 36 ¸ 12 = 3

Median: arranging the numbers in order we have

1 1 2 2 2 2 <- middle -> 3 4 4 4 4 7

Since there is an even number of data points, no one number is in the middle and we have to find the average of the two middle numbers. Therefore the median is (2 + 3)/2 = 2.5.

Mode: the most common numbers are 2 and 4 (there are four each) so there are two modes, 2 and 4.

A common test question requires finding a missing number in a set of numbers given the other numbers and the mean. For example, if there are 12 numbers and you know 11 of them as well as the average of the 12 numbers, you can find the twelfth number.

Example: The average of 12 numbers is 3. Given the following 11 numbers,

1 1 2 2 2 2 3 4 4 4 7

find the missing number.

Let x the missing number. Then

(1 + 1 + 2 + 2 + 2 + 2 + 3 + 4 + 4 + 4 + 4 + 7 + x)/12 = 3

(33 + x)/12 = 3

33 + x = 36

x = 3

Example:

The average of four quiz grades is 72%. If the student wishes to have at least an 80% quiz average and if there are two more quizzes to be taken, what must the average of the last two quizzes be?

A. 76

B. 84

C. 88

D. 92

E. 96

Answer: E

Solution: If the student has taken four quizzes and the average is 72 then the total quiz score is 72x4=288. If the student wants an 80 average on six quizzes then he or she must have 80x6=480 total points. Therefore, the student needs an additional 192 points (480-288) on two quizzes. Since 192/2=96 the student needs an average of 96 points on the last two quizzes.

Extra: An easy way to see this is to think of each of the first four quizzes as being equal to 72 points. (Saying an average is 72 is like saying each grade is 72, although none of the grades might actually be equal to 72.) Let x be the average of the last two grades, which is like each grade is x. Then

(72+72+72+72+x+x) ¸ 6 = 80

(72+72+72+72+x+x) = 480

288 +2x = 480

2x = 192

x = 96

Example:

The table below gives the ages of 5 people:

Person	Betty	Ralph	Jesse	Rea	Eric
Age	63	72	31	31	9

Find the average age (arithmetic mean).

A. 31

B. 40.5

C. 41.2

D. 51.5

E. 206

Answer: C

Solution: Add the ages and divide by the number of people.

(63+72+31+31+9)/5=206/5=41.2

Similar problem: The data might have been presented in the following manner:

Age	Number of people
9	1
31	2
63	1
72	1

Here every number is not weighted equally. In this case 31 must be counted twice. We will be finding a weighted average, wherein each data point is weighted by the number of times it is counted.

Weighted average: [1(9)+2(31)+1(63)+1(72)]/5 = 206/5=41.2