3STDEV1 Engineering Stat

Engineering Statistics

MEASURES OF SPREADCHAP. 3.3 – 3.5

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VARIATION• RANGE:

• PERCENTILES AND QUARTILES:Q1

• INTERQUARTILE RANGE: Q3 – Q1

• BOX AND WHISKER PLOTS

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Variance: reflects the values of all the measurements

• Population variance:

• Sample variance 2

2 ( )1

x xs

n

22 ( )x

N

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The Standard Deviation• Compares the variability of several

distributions• Makes a statement about the general shape

of a distribution• Square root of variance

1

)(1

1)(

222

n

fmn

fm

nxmf

s

5

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Examples of finding standard deviation

• A professor wants to report the results of a test taken by 100 students. Find the mean, median, mode and standard deviation.

• x = 73.98 kurtosis = 0.3936606• Median = 81 skewness = -1.073098• Mode = 84 range = 89 • s = 21.502163 n =100 Describe the original data

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Examples:

• Standard Deviation of Income Distribution • Explorer• Defining the Risk factor using standard

deviation

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Two mutual funds in dollars:• Fund A: 8.3, -6.2, 20.9, -2.7, 33.6, 42.9,

24.4, 5.2, 3.1, 30.05 x = 16, s = 16.74• Fund B: 12.1, -2.8, 6.4, 12.2, 27.8, 25.3,

18.2, 10.7, -1.3, 11.4 x = 12, s = 9.969.Which is riskier?

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Looking beyond the average

• How typical is the average value of all the measurements in the data set?

• How spread out are the measurements about the average value?

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Examples of finding standard deviation

• A professor wants to report the results of a test taken by 100 students. Find the mean, median, mode and standard deviation.

• x =73.98 kurtosis = 0.3936606• Median = 81 skewness = -

1.073098• Mode = 84 range = 89 • s = 21.502163 n =100Describe the original data

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Two mutual funds in dollars:• Fund A: 8.3, -6.2, 20.9, -2.7, 33.6, 42.9,

24.4, 5.2, 3.1, 30.05 x = 16, s = 16.74• Fund B: 12.1, -2.8, 6.4, 12.2, 27.8, 25.3,

18.2, 10.7, -1.3, 11.4 x = 12, s = 9.969.Which is riskier? Defining the Risk factor using standard

deviation

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Find the combination of a lock!

• The combination of a lock is given by the following distribution: 14, 5, 6, 11, 13, 8, 3, and 4.

• Turn left the mean of the distribution, turn right the value of the median, turn left the largest positive integer in the standard deviation and turn right 0.5 plus half the range.

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Examples:

• Standard Deviation of Income Distribution • http://wbln0018.worldbank.org/psd/

compete.nsf/66ce914fdee59ef585256490006019d1?OpenView

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Unimodal Distributions• Skewness: extent of departure from

symmetry: a3 = 0 is symmetrical

• Kurtosis: extent of peakedness of a bell-shaped distribution in comparison to the normal distribution: a4 = 4.00 for a normal curve 4

4 4

1 ( )x xna =

s

3

3 3

1 ( )x xna =

s

3

3)(1

s

xmfn

4

4)(1

s

xmfn

16

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Coefficient of Variation

• a set of measurements • standard deviation divided by the mean

value • CV = s/x• a proportionate measure of variation

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Z-scores

• The distance between a measurement and its mean in standard deviation units

•

• Example:

x x xz or zs

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Chebyshev’s Theorem

• applies to any smoothed distribution or histogram

the percentage of the total area under any smoothed distribution is at least 100(1-1/k2)% (k standard deviations)

(x – 2s, x + 2s) contains at least 75% of the measurements

(x – 3s, x +3s) contains at least 89% of the measurements

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The Empirical Rule

• If a sample of measurements has a mound-shape distribution, the interval

(x – s, x + s) contains approximately 68% of the measurements

(x – 2s, x + 2s) contains approximately 95% of the measurements

(x – 3s, x +3s) contains approximately 99.7% of the measurements

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The Empirical Rule

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The Empirical Rule

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Chebychev and Empirical Rules Example:

• A pharmaceutical company manufactures vitamin pills which contain an average of 507 grams of vitamin C with a standard deviation of 3 grams.

Using Chebychev's rule, we know that at least 75% of the vitamin pills are within k = 2 standard deviations of the mean. That is, at least 75% of the vitamin pills will have between 501 and 513 grams of vitamin C.

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Chebychev and Empirical Rules Example cont’d.

• If the distribution of vitamin C amounts in the previous example is bell shaped, then we can get even more precise results by using the Empirical rule.

Under these conditions, approximately 68% of the vitamin pills have a vitamin C content in the interval [507-3, 507+3]=[504,510], 95% are in the interval [507-2(3), 507+2(3)]=[501,513], and 99.7% are in the interval [507-3(3), 507+3(3)]=[498,516].

• NOTE: Chebychev's rule gives only a minimum proportion of observations which lie within k standard deviations of the mean.

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Z-Score Problems:1. Scores on intelligence tests (IQs) are normally

distributed in children. IQs from the Wechsler intelligence tests are known to have means of 100 and standard deviations of 15. In almost all the states in the United States (Pennsylvania and Nebraska are exceptions) children can be labeled as mentally retarded if their IQ falls to 70 points or below. What is the maximum z score one could obtain on an intelligence test and still be considered to be mentally retarded?

2. If gifted children need IQs of 130 and above, what is the minimum z score on an intelligence test that a gifted child can obtain?

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Z-Score Problems cont’d.3. Using the normal curve Empirical Rule, what

percentage of the children in the United States are thought to be neither retarded nor gifted?

4. What percent have IQs between ±2 z?

5. What percentage of the children would be thought to be retarded?

6. What percentage of the children would be expected to be gifted in the typical school system?

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• Answers #3 & 4: The percentage of children who are neither retarded or gifted and have IQs between ±2 z would equal 95% using the empirical rule.

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• Answer #5: Given a normal distribution, you know that 50% of the children have IQs below the mean. If 95% of the children have IQs between ±2 z, then half that percentage (the curve is symmetrical) would be below the mean and a -2 z. Half of 95% is 47.5%. Thus the percentage of children thought to be retarded would be 50% - 47.5% = 2.5%. The Figure illustrates the answer to this problem.

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• Answer #6: This is really the same problem as Question 2, except you are dealing with the upper tail in the normal distribution. The answer is the same, 2.5%, and the Figure on the left illustrates this solution.

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MEASURES OF SPREAD

• Variance• Standard Deviation• Z-Score• Chebyshev and Empirical Rules

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Exercises:

• p. 99 # 3.37 • p. 107 # 3.55

• Homework: pp. 110 – 116 # 5, 7, 9, 11, 13, 17, 19