#### Mean

TheÂ ** arithmetic mean**Â is a number often used to describe the “average” value of a data set. The arithmetic mean is defined as:

mean = \dfrac{sum \,of\, the \,values}{number \,of \,values}

#### Median

TheÂ ** median**Â is the “middle” value. To find the median, arrange the values in numerical order. If the number of values is odd, the median is the middle value. If the number of values is even, the median is the mean of the two middle values.

#### Range

TheÂ ** range**Â is a number that describes the spread or dispersion of a data set. The range is the difference between the largest and smallest values.

range = (greatest value) â€“ (least value)

## Example

Find the arithmetic mean, median and range.

(a)Â 6, 9, 11, 13, 18, 21

(b)Â 2, 45, 19, 3, 6

### Solution

**(a) Â Â Â Â Â Â **mean = \dfrac{6 + 9 + 11 + 13 + 18 + 21}{6}

= 13

There are an odd number of values, so the median is the mean of the two middle numbers.

median = 11 + \dfrac{13}{2} = 12

range = 21 â€“ 6 = 15

**(b) Â Â Â Â Â Â **

mean = \dfrac{2 + 45 + 19 + 3 + 6}{5}

= 15

Arranging the values in numerical order makes it easier to see the median and range. 2, 3, 6, 19, 45

The number of values is odd, so the median is the middle number.

median = 6

range = 45 â€“ 2 = 43

## Comparing Data

Every set of numbers has a mean, median and range. These three values can be used to compare data by seeing how much the numbers differ.

## Example

A politician announces at a press conference that the average yearly salary per person in the state is now about $80,000.

After her announcement, three groups of 5 people were surveyed about their salaries. Is the claim about average salaries reasonable for each group?

(I)Â $80,000;Â $80,000;Â $80,000;Â $80,000;Â $80,000

(II)Â $71,000;Â $74,000;Â $82,500;Â $82,500;Â $90,000

(III)Â $15,000;Â $27,000;Â $33,000;Â $100,000;Â $225,000

### Solution

To solve this problem, compare the mean, median and range of each data set.

In (I), the mean and median are $80,000. The range is $0. The claim is accurate.

In (II), the mean is $80,000. The median is $82,500. The range is $19,000. So the mean and median support the claim. The range is somewhat high, showing the data is spread out. The claim is still somewhat reasonable.

In (III), the mean is $80,000. The median is $30,000. The range is $210,000. The claim is not reasonable for this data set.

## Standard Deviation

TheÂ ** standard deviation** compares data by looking at how much the numbers in a set differ, or deviate, from the mean. The greater the difference, the greater the standard deviation.

A small standard deviation means the numbers are all clustered around the mean.Â A large deviation means the numbers are spread out from the mean.

## 800Score Tip:

The GRE does not require you to calculate the Standard Deviation or know its formula. However, you need to understand what standard deviation is, how it works, what numbers it compares and what it represents.

The formula for standard deviation is

*sd* = \sqrt{\dfrac{sum \,of\, (each \,value \,-\, mean)^{\displaystyle{2}}}{number \,of \,values}}

Calculating the standard deviation requires knowing all the numbers in the set.

## Example

Compare the 3 sets of numbers. Which set has the greatest standard deviation?

(I) 4, 4, 4, 4, 4

(II) 1, 3, 4, 5, 7

(III) 0, 0, 0, 10, 10

### Solution

Start by just looking at the sets.

Notice that all lists have the same average, 4. Compare the differences from the mean. Which list has the most deviation from 4?

(I) | (II) | (III) | |

4, 4, 4, 4, 4 | 1, 3, 4, 5, 7 | 0, 0, 0, 10, 10 | |

mean | 4 | 4 | 4 |

differences from mean | 0, 0, 0, 0, 0 | 3, 1, 0, 1, 3 | 4, 4, 4, 6, 6 |

deviation | none | moderate | large |

So Set (III) has the greatest standard deviation.

Answering this question does not require calculating the standard deviation.

But the actual standard deviations are:

(I) | (II) | (III) | |

standard deviation | 0 | 2 | \sqrt{24} â‰ˆ 5 |

## Example

Compare the 3 sets of numbers. Which set has the greatest standard deviation?

(I) 16, 17, 18, 19, 20

(II) 4, 6, 8, 10, 12

(III) 48, 50, 52, 54, 56

### Solution

The distances between adjacent numbers in Set (I) are 1. The distances between adjacent numbers in Set (II) are 2. So Set (I) cannot have the greatest standard deviation.

The mean of Set (II) is 8. The means of Set (III) is 52. While the mean of Set (III) is greater than the mean of Set (II), the numbers of Set (III) are the same distance from its mean of 52 as the numbers of Set (II) are from its mean of 8. They are both lists of consecutive even numbers, so they have the same dispersion.

This is a trick question. Set (II) and Set (III) have the same standard deviation.

## Example

(a)If 10 is added to each number in a set, what is the change in the mean? What is the change in the standard deviation?

(b)If each number in a set is multiplied by 10, what is the change in the mean? What is the change in the standard deviation?

### Solution

**(a)** Since each value increases by 10, the mean increases by 10.

Adding 10 to each number does not change the distances between the numbers and the mean. Since the standard deviation is dependent on those distances, the standard deviation does not change.

If you are not sure about the answer, you can Plug In some numbers and try it.

The mean of {1, 2, 3} is 2. The mean of {11, 12, 13} is 12. The mean increased by 10.

The distances between {1, 2, 3} and their mean are 1. The distances between {11, 12, 13} and their mean are also 1. The standard deviation did not change.

**(b)** Since each value is multiplied by 10, the mean is multiplied by 10.

Multiplying each number by 10 changes the distances between the numbers and the mean by a factor of 10. Since the standard deviation is dependent on those distances, the standard deviation changes by a factor of 10.

If you are not sure about the answer, you can Plug In some numbers and try it.

The mean of {1, 2, 3} is 2. The mean of {10, 20, 30} is 20. The mean increased by a factor of 10.

The distances between {1, 2, 3} and their mean are 1. The distances between {10, 20, 30} and their mean are 10. Since the distances changed by a factor of 10, the standard deviation also changed by a factor of 10.

## Example

IfÂ

aÂ is a positive number, then which set has the greatest standard deviation?(I)Â 0, 0, 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(II)Â Âa,Âa,ÂaÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(III)Â 0, 0,ÂaA. I

B. II

C. III

D. All are equal.

E. It cannot be determined.

### Solution

The standard deviation of Set (I) is 0, because all the elements are equal.

The standard deviation of Set (III) is 0, because all elements are equal.

The standard deviation of Set (III) is greater than 0 because *a* is a positive number, so set (III) has the greatest standard deviation.

The correct answer is choice (C).

## Example

The standard deviation of setÂ

Q, which consists of 4 integers, is 40. If every element of the setQÂ is increased by its standard deviation, what will be the standard deviation of the new set?A. 20

B. 40

C. 40 + 2\sqrt{10}

D. 80

E. 1600

### Solution

A standard deviation does not change if the same number is added to each value in the set. In this case the standard deviation (40) is added to each value, therefore the standard deviation of the newly created set equals the standard deviation of the original set.

The correct answer is choice (B).

#### Video Quiz

#### Standard Deviation Introduction

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1 question with a video explanation

100 seconds per question