Today s plan: Section 4.1.4: Dispersion: Five-Number summary and Standard Deviation.
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1 1 Today s plan: Section 4.1.4: Dispersion: Five-Number summary and Standard Deviation.
2 2 Once we know the central location of a data set, we want to know how close things are to the center.
3 2 Once we know the central location of a data set, we want to know how close things are to the center. We ll see two ways to measure dispersion of a data set.
4 3 five-number summary (goes with the median)
5 3 five-number summary (goes with the median) standard deviation (goes with the mean)
6 4 Five-Number Summary
7 5 Five-number Summary: 1. Min 2. Lower Quartile 3. Median 4. Upper Quartile 5. Max
8 6 Definition The Min is the smallest value in the whole data set.
9 6 Definition The Min is the smallest value in the whole data set. The Max is the largest value in the whole data set.
10 6 Definition The Min is the smallest value in the whole data set. The Max is the largest value in the whole data set. The Lower Quartile is the median of the lower half.
11 6 Definition The Min is the smallest value in the whole data set. The Max is the largest value in the whole data set. The Lower Quartile is the median of the lower half. The Upper Quartile is the median of the upper half.
12 7 Example The appraisals of the 10 houses are: [$75K, $96K, $107K, $110K, $110K, $118K, $130K, $135K, $150K, $520K ]
13 7 Example The appraisals of the 10 houses are: [$75K, $96K, $107K, $110K, $110K, $118K, $130K, $135K, $150K, $520K ] Find the five-number summary.
14 8 Solution We already found: the median, Med = $114K
15 8 Solution We already found: the median, Med = $114K the lower half, [$75K, $96K, $107K, $110K, $110K ]
16 8 Solution We already found: the median, Med = $114K the lower half, [$75K, $96K, $107K, $110K, $110K ] the upper half [$118K, $130K, $135K, $150K, $520K ]
17 8 Solution We already found: the median, Med = $114K the lower half, [$75K, $96K, $107K, $110K, $110K ] the upper half [$118K, $130K, $135K, $150K, $520K ] Since each half has size 5, their respective medians will be in the 3rd location.
18 9 Solution Thus the lower quartile is Q1 = $107K
19 9 Solution Thus the lower quartile is Q1 = $107K the upper quartile is Q3 = $135K
20 9 Solution Thus the lower quartile is Q1 = $107K the upper quartile is Q3 = $135K the lowest value is Min = $75K
21 9 Solution Thus the lower quartile is Q1 = $107K the upper quartile is Q3 = $135K the lowest value is Min = $75K the highest value is Max = $520K
22 9 Solution Thus the lower quartile is Q1 = $107K the upper quartile is Q3 = $135K the lowest value is Min = $75K the highest value is Max = $520K So the five-number summary is: [Min = $75K, Q1 = $107K, Med = $114K, Q3 = $135K, Max = $520K ].
23 10 The five-number summary can be visualized with a boxplot diagram, or box-and-whiskers diagram.
24 Min Q1Med Q3 Max \\
25 12 The box goes from the lower quartile to the upper quartile, with a mark at the median.
26 12 The box goes from the lower quartile to the upper quartile, with a mark at the median. Two whiskers extend from the box to the Min and Max.
27 13 Remarks: the left whisker spans the bottom 25%
28 13 Remarks: the left whisker spans the bottom 25% the box spans the middle 50%
29 13 Remarks: the left whisker spans the bottom 25% the box spans the middle 50% the right whisker spans the top 25%
30 13 Remarks: the left whisker spans the bottom 25% the box spans the middle 50% the right whisker spans the top 25% each half of the box spans 25%
31 14 Example The ages of the police officers in the Clearview Police Department are Age Freq
32 14 Example The ages of the police officers in the Clearview Police Department are Age Freq Find the five-number summary and draw the boxplot.
33 15 Age Freq Cum. Freq
34 16 The size is n = 41, so the median is in location
35 16 The size is n = 41, so the median is in location = 21. 2
36 16 The size is n = 41, so the median is in location = The lower half has size 20, so the lower quartile is the average of the values at locations 10 and 11: Q1 = =
37 17 The upper half also has size 20, so the upper quartile is the average of the values at locations 10 and 11 of the upper half.
38 17 The upper half also has size 20, so the upper quartile is the average of the values at locations 10 and 11 of the upper half. Since the median is at location 21, the third quartile is the average of the values at locations 31 and 32 of the whole data set: Q3 = = 32
39 18 Five-number summary: [Min = 22, Q1 = 26.5, Med = 29, Q3 = 32, Max = 39] Min Q1 Med Q3 Max
40 19 Remark: Outliers can be drawn separated from the rest of the data set.
41 20 Example The appraisals of the 10 houses are: [$75K, $96K, $107K, $110K, $110K, $118K, $130K, $135K, $150K, $520K ]
42 20 Example The appraisals of the 10 houses are: [$75K, $96K, $107K, $110K, $110K, $118K, $130K, $135K, $150K, $520K ] Find the five-number summary with outliers separated.
43 Min Q1Med Q3 Max \\
44 22 Boxplots and five-number summaries are useful when comparing two data sets.
45 23 Example Waiting times at two car washes: Acme Car Wash: [Min = 1, Q1 = 5, Med = 8, Q3 = 9, Max = 12] Kleen Car Wash: [Min = 3, Q1 = 4, Med = 5, Q3 = 8, Max = 20] (Times are in minutes.)
46 24 Example Draw the boxplots together, and compare them.
47 25 Solution Here are the boxplots: Acme Kleen
48 26 Solution The Min and Max tell us:
49 26 Solution The Min and Max tell us: everyone at Kleen has to wait at least 3 minutes, and some people have a very long wait.
50 26 Solution The Min and Max tell us: everyone at Kleen has to wait at least 3 minutes, and some people have a very long wait. at Acme, some have a tiny wait and everyone gets started in 12 minutes.
51 26 Solution The Min and Max tell us: everyone at Kleen has to wait at least 3 minutes, and some people have a very long wait. at Acme, some have a tiny wait and everyone gets started in 12 minutes. Acme seems better.
52 27 Solution But, the Median tells us: half of the customers of Acme wait 8 minutes for service
53 27 Solution But, the Median tells us: half of the customers of Acme wait 8 minutes for service at Kleen half of them start in 5 minutes
54 27 Solution But, the Median tells us: half of the customers of Acme wait 8 minutes for service at Kleen half of them start in 5 minutes Now Kleen seems better.
55 28 Which is better? There s no simple answer
56 28 Which is better? There s no simple answer If you don t mind waiting a little, Acme is better, since there are no long waits.
57 28 Which is better? There s no simple answer If you don t mind waiting a little, Acme is better, since there are no long waits. If you re willing to risk a long wait, in hope of a really short wait, Kleen is better.
58 29 Standard Deviation
59 30 When using the mean to measure the center, we use the standard deviation to measure dispersion.
60 30 When using the mean to measure the center, we use the standard deviation to measure dispersion. Think of standard deviation as measuring how far from the average the data points tend to be.
61 31 (Wrong way:)
62 31 (Wrong way:) 1. take the deviation of each data point from the average
63 31 (Wrong way:) 1. take the deviation of each data point from the average 2. average those deviations
64 31 (Wrong way:) 1. take the deviation of each data point from the average 2. average those deviations The deviation of a point x i from the average x is just x i x
65 32 (Wrong way:)
66 32 (Wrong way:) Example Weekly Sales of Home Town Pharmacy: S M T W R F S $2,548, $1,225, $1,732, $1,871, $975, $2,218, $1,339. Find the average of x i x.
67 32 (Wrong way:) Example Weekly Sales of Home Town Pharmacy: S M T W R F S $2,548, $1,225, $1,732, $1,871, $975, $2,218, $1,339. Find the average of x i x. We have already found the average: x =
68 33 (Wrong way:) Here are deviations x i x: Day x i (sales) x i x (deviation) Sunday 2, Monday 1, Tuesday 1, Wednesday 1, Thursday Friday 2, Saturday 1, Total 11, Average 1,
69 34 (Wrong way:) Deviations are like distances, but with a sign
70 34 (Wrong way:) Deviations are like distances, but with a sign Positive deviation x i is to the right of x
71 34 (Wrong way:) Deviations are like distances, but with a sign Positive deviation x i is to the right of x Negative deviation x i is to the left of x
72 35 (Wrong way:) The average of those deviations: = 0.00
73 35 (Wrong way:) The average of those deviations: This is going to happen with any data set! Average deviation from the mean is a useless measure of dispersion. = 0.00
74 36 (Right way:) However, if we square all deviations, they will turn all positive
75 36 (Right way:) However, if we square all deviations, they will turn all positive We can then average those squared deviations
76 36 (Right way:) However, if we square all deviations, they will turn all positive We can then average those squared deviations that is called the variance
77 37 Definition The variance var(x) of a data set x is the average of the squared deviations from the mean x: var(x) = 1 (xi x) 2 n
78 38 To compensate for the squaring, we take the square root of the variance.
79 38 To compensate for the squaring, we take the square root of the variance. Definition The standard deviation is σ(x) = var(x)
80 39 Example Find the variance and standard deviation for the Home Town Pharmacy daily sales data set.
81 40 Day x (sales) x x (x x) 2 Sunday 2, Monday 1, Tuesday 1, Wednesday 1, Thursday Friday 2, Saturday 1, Total 11, Average 1,
82 41 the variance is var(x) =
83 41 the variance is var(x) = the standard deviation is σ(x) = =
84 42 What if we start with a frequency table or a histogram?
85 43 Example Find the standard deviation for the Math 109 quizzes score freq cum fr
86 44 Solution We computed the average µ = 14.64
87 44 Solution We computed the average µ = For convenience turn the frequency table into a vertical table
88 45 x f x f (x µ) (x µ) 2 (x µ) 2 f Tot Ave
89 46 So the standard deviation is σ = = 3.35.
90 47 To find the Standard Deviation σ 1. Compute the deviations x i µ. 2. Square the deviations (x i µ) Average the squared deviations to the variance (xi µ) 2 var =. n 4. Take the square root of the variance σ = var.
91 48 Question What does standard deviation mean in practice?
92 49 In the previous example: The average is µ = the standard deviation is σ = 3.35
93 50 How many data points are within one standard deviation of the average?
94 50 How many data points are within one standard deviation of the average? µ σ = and µ + σ = 17.99
95 50 How many data points are within one standard deviation of the average? µ σ = and µ + σ = Between these two values there are a total of = 62 data points (out of 95), i.e., about two thirds.
96 51 For nice data sets, about 2 of the 3 data set is located within one standard deviation of the average.
97 51 For nice data sets, about 2 of the 3 data set is located within one standard deviation of the average. if σ is small, the data points are crowded close to µ
98 51 For nice data sets, about 2 of the 3 data set is located within one standard deviation of the average. if σ is small, the data points are crowded close to µ if σ is large, the data points are scattered.
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