• According to researchers, the average American guy is 31 years old, 5 feet 10 inches, 172 pounds, works 6.1 hours daily, and sleeps 7.7 hours.

• These numbers represent what is “average” or “typical” of American men.

• In statistics, such values are known as measures of central tendency because they are generally located toward the center of a distribution.

Measures of Central TendencyMeasures of Central Tendency

xx

n

Most frequently used measure of central tendency

Strongly influenced by outliers - very large or very small values

MeanArithmetic average

Sum of all data values divided by the number of data values in an array

x

Measures of Central TendencyMeasures of Central Tendency

xx

n

48, 63, 62, 49, 58, 2, 63, 5, 60, 59, 55Determine the mean value of

( 48 63 62 49 58 2 63 5 60 59 55)x

11

524x

11

x 47.64

Measures of Central TendencyMeasures of Central TendencyMedian

• The middle number• Data values must be ordered from lowest to

highest• If you have an odd number of data then the

median is the value in the middle of the set• If you have an even number of data then the

median is the average between the two middle values in the set

• Useful in situations with skewed data and outliers

Measures of Central TendencyMeasures of Central TendencyDetermine the median value of

Organize the data array from lowest to highest value.

59, 60, 62, 63, 63

48, 63, 62, 49, 58, 2, 63, 5, 60, 59, 55

Select the data value that splits the data set evenly.

2, 5, 48, 49, 55, 58,

Median = 58.5

What if the data array had an even number of values?

60, 62, 63, 635, 48, 49, 55, 58, 59,

Median = 58

Position of the Median

If n data items are arranged in order, from smallest to largest, the median is the

value in the (n + 1) / 2 position

Measures of central tendencyMeasures of central tendencyMode

• Most frequently occurring response within a data array

• May not exist at all• If no numbers repeat then the mode = 0

• Bimodal & multimodal

Measures of Central TendencyMeasures of Central TendencyDetermine the mode of

48, 63, 62, 49, 58, 2, 63, 5, 60, 59, 55Mode = 63

Determine the mode of

48, 63, 62, 59, 58, 2, 63, 5, 60, 59, 55Mode = 63 & 59 Bimodal

Determine the mode of

48, 63, 62, 59, 48, 2, 63, 5, 60, 59, 55Mode = 63, 59, & 48 Multimodal

Suppose your six exam grades in a class are:

52, 69, 75, 86, 86, 92.

Compute your final grade (90-100 = A, 80-89 = B, 70-79 = C, 60-69 = D, below 60 = F) using the:

a. mean b. median c. mode

d. Find the median position

Meaures of DispersionMeaures of Dispersion

Range

Standard Deviation

Variance

Measure of data scatter

Difference between the lowest and highest data value

Square root of the variance

Average of squared differences between each data value and the mean

RangeRange

R 63 2

Calculate by subtracting the lowest value from the highest value.

R h l

2, 5, 48, 49, 55, 58, 59, 60, 62, 63, 63

Calculate the range for the data array.

R h l

R 61

Standard DeviationStandard Deviation 2x xs

( N 1)

1. Calculate the mean .

2. Subtract the mean from each value.

3. Square each difference.

4. Sum all squared differences.

5. Divide the summation by the number of values in the array minus 1.

6. Calculate the square root of the product.

x

Standard DeviationStandard Deviation 2x xs

( N 1)

2, 5, 48, 49, 55, 58, 59, 60, 62, 63, 63

Calculate the standard deviation for the data array.

x

x

n

52411

1. 47.64

2. 2 – 47.64 = -45.64

5 – 47.64 = -42.64

48 – 47.64 = 0.36

49 – 47.64 = 1.36

55 – 47.64 = 7.36

58 – 47.64 = 10.36

59 – 47.64 = 11.36

60 – 47.64 = 12.36

62 – 47.64 = 14.36

63 – 47.64 = 15.36

63 – 47.64 = 15.36

x x

Standard DeviationStandard Deviation 2x xs

( N 1)

2, 5, 48, 49, 55, 58, 59, 60, 62, 63, 63

Calculate the standard deviation for the data array.

3.

(-45.64)2 = 2083.01

(-42.64)2 = 1818.17

0.362 = 0.13

1.362 = 1.85

7.362 = 54.17

10.362 = 107.33

11.362 = 129.05

12.362 = 152.77

14.362 = 206.21

15.362 = 235.93

15.362 = 235.93

2x x

Standard DeviationStandard Deviation 2x xs

( N 1)

2, 5, 48, 49, 55, 58, 59, 60, 62, 63, 63

Calculate the standard deviation for the data array.

4.

2083.01 + 1818.17 + 0.13 + 1.85 + 54.17 + 107.33 + 129.05 + 152.77 + 206.21 + 235.93 + 235.93

2x x

= 5,024.555.( N 1)

11-1 = 10

6. 2( 1

x x

N )

5,024.5510

502.46

7. 2x xs

( N 1)

502.46S = 22.42

Standard DeviationStandard Deviation

• The standard deviation measures the spread of the distribution or the dispersion the distribution has.

• The larger the standard deviation, the more spread out the distribution is.

• The smaller the standard deviation, the less spread the distribution has.

VarianceVariance 22x x

s( N 1)

Steps 1 – 5 of the standard deviation:

1.Calculate the mean.

2.Subtract the mean from each value.

3.Square each difference.

4.Sum all squared differences.

5.Divide the summation by the number of values in the array minus 1.

VarianceVariance

2 5024.55s

(50

1 )46

02.

2, 5, 48, 49, 55, 58, 59, 60, 62, 63, 63

Calculate the variance for the data array.

22x x

s( N 1)

Variance (sVariance (s22))

• Mathematically expressing the degree of variation of data from the mean

• A large variance means that the individual data of the sample deviate a lot from the mean.

• A small variance indicates the data deviate little from the mean