according to researchers, the average american guy is 31 years old, 5 feet 10 inches, 172 pounds,...
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• 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