#set document(title: "3.2 Measures of Spread", author: "Kathryn Kozak")
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== 3.2#h(0.6em)Measures of Spread


    Variability is an important idea in statistics. If you were to measure the height of everyone in your classroom, every observation gives you a different value. That means not every student has the same height. Thus there is variability in people’s heights. If you were to take a sample of the income level of people in a town, every sample gives you different information. There is variability between samples too. Variability describes how the data are spread out. If the data are very close to each other, then there is low variability. If the data are very spread out, then there is high variability. How do you measure variability? It would be good to have a number that measures it. This section will describe some of the different measures of variability, also known as variation.


    In Example #math.equation(block: false, alt: "1")[$1$], the average weight of a cat was calculated to be 8.02 pounds. How much does this tell you about the weight of all cats? Can you tell if most of the weights were close to 8.02 or were the weights really spread out? What are the highest weight and the lowest weight? All you know is that the center of the weights is 8.02 pounds. You need more information.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The #strong[range]of a set of data is the difference between the highest and the lowest data values (or maximum and minimum values).


      #math.equation(block: true, alt: "Range equals highest value minus lowest value; equals maximum value minus minimum value")[$"Range" = "highest value" − "lowest value" \ = "maximum value" − "minimum value"$]
]


    #examplebox("Example 1")[Finding the Range][
Look at the following three sets of data. Find the range of each of these.


        + #math.equation(block: false, alt: "10 , 20 , 30 , 40 , 50")[$10 , 20 , 30 , 40 , 50$]
+ #math.equation(block: false, alt: "10 , 29 , 30 , 31 , 50")[$10 , 29 , 30 , 31 , 50$]
+ #math.equation(block: false, alt: "28 , 29 , 30 , 31 , 32")[$28 , 29 , 30 , 31 , 32$]


      
      #solutionbox[
a.


        #figure(figph[A dot plot with one dot each at 10, 20, 30, 40, and 50, showing mean and median of 30 and range of 40.], alt: "A dot plot with one dot each at 10, 20, 30, 40, and 50, showing mean and median of 30 and range of 40.", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: Dot Plot for Example #math.equation(block: false, alt: "1")[$1$]a])


        b.


        #figure(figph[A dotplot shows five values at 10, 28, 30, 32, and 50, with mean and median 30 and range 40.], alt: "A dotplot shows five values at 10, 28, 30, 32, and 50, with mean and median 30 and range 40.", caption: [Figure #math.equation(block: false, alt: "2")[$2$]: Dot Plot for Example #math.equation(block: false, alt: "1")[$1$]b])


        c.


        #figure(figph[A dot plot showing five values from 28 to 32, centered at 30.], alt: "A dot plot showing five values from 28 to 32, centered at 30.", caption: [Figure #math.equation(block: false, alt: "3")[$3$]: Dot Plot for Example #math.equation(block: false, alt: "1")[$1$]])


        Based on the mean, median, and range in Example #math.equation(block: false, alt: "1")[$1$], the first two distributions are the same, but you can see from the graphs that they are different. In Example #math.equation(block: false, alt: "1")[$1$]a the data are spread out equally. In Example #math.equation(block: false, alt: "1")[$1$]b the data has a clump in the middle and a single value at each end. The mean and median are the same for Example #math.equation(block: false, alt: "1")[$1$]c but the range is very different. All the data is clumped together in the middle.
]
]


    The range doesn’t really provide a very accurate picture of the variability. A better way to describe how the data is spread out is needed. Instead of looking at the distance the highest value is from the lowest how about looking at the distance each value is from the mean. This distance is called the #strong[deviation].


    #examplebox("Example 2")[Finding the Deviations][
Suppose a vet wants to analyze the weights of cats. The weights (in pounds) of five cats are 6.8, 8.2, 7.5, 9.4, and 8.2. Find the deviation for each of the data values.


      
      #solutionbox[
Variable: #math.equation(block: false, alt: "x equals")[$x =$] weight of a cat


        The mean for this data set is #math.equation(block: false, alt: "x ― equals 8.02")[$limits(x)^(―) = 8.02$] pounds.


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x minus x ―")[$x − limits(x)^(―)$]]),
  [6.8], [6.8-8.02 = -1.22],
  [8.2], [8.2-8.02=0.18],
  [7.5], [7.5-8.02=-0.52],
  [9.4], [9.4-8.02=1.38],
  [8.2], [8.2-8.02=0.18],
))


        Now you might want to average the deviation, so you need to add the deviations together.


        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x minus x ―")[$x − limits(x)^(―)$]]),
  [6.8], [6.8-8.02 = -1.22],
  [8.2], [8.2-8.02=0.18],
  [7.5], [7.5-8.02=-0.52],
  [9.4], [9.4-8.02=1.38],
  [8.2], [8.2-8.02=0.18],
  [Total], [0],
))


        This can’t be right. The average distance from the mean cannot be 0. The reason it adds to 0 is because there are some positive and negative values. You need to get rid of the negative signs. How can you do that? You could square each deviation.


        #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x minus x ―")[$x − limits(x)^(―)$]], [#math.equation(block: false, alt: "open parenthesis x minus x ― close parenthesis squared")[$( x − limits(x)^(―) )^(2)$]]),
  [6.8], [6.8-8.02 = -1.22], [1.4884],
  [8.2], [8.2-8.02=0.18], [0.0324],
  [7.5], [7.5-8.02=-0.52], [0.2704],
  [9.4], [9.4-8.02=1.38], [1.9044],
  [8.2], [8.2-8.02=0.18], [0.0324],
  [Total], [0], [3.728],
))


        Now average the total of the squared deviations. The only thing is that in statistics there is a strange average here. Instead of dividing by the number of data values you divide by the number of data values minus 1. In this case you would have


        #math.equation(block: true, alt: "s squared equals the fraction 3.728 over 5 minus 1 equals the fraction 3.728 over 4 equals 0.932 pounds squared")[$s^(2) = frac(3.728, 5 − 1) = frac(3.728, 4) = 0.932 attach(" pounds ", t: 2)$]


        Notice that this is denoted as #math.equation(block: false, alt: "s squared")[$s^(2)$]. This is called the variance and it is a measure of the average squared distance from the mean. If you now take the square root, you will get the average distance from the mean. This is called the standard deviation, and is denoted with the letter #math.equation(block: false, alt: "s")[$s$].


        #math.equation(block: false, alt: "s equals the square root of .932 approximately equals 0.965")[$s = sqrt(".932") ≈ 0.965$] pounds
]
]


    The standard deviation is the average (mean) distance from a data point to the mean. It can be thought of as how much a typical data point differs from the mean.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Sample Variance]


      The #strong[sample variance]formula:


      #math.equation(block: true, alt: "s squared equals the fraction ∑ open parenthesis x minus x ― close parenthesis squared over n minus 1")[$s^(2) = frac(∑ ( x − limits(x)^(―) )^(2), n − 1)$]


      where #math.equation(block: false, alt: "x ―")[$limits(x)^(―)$] is the sample mean, #math.equation(block: false, alt: "n")[$n$] is the sample size, and #math.equation(block: false, alt: "∑")[$∑$] means to find the sum.
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Sample Standard Deviation]


      The #strong[sample standard deviation]formula:


      #math.equation(block: true, alt: "s equals the square root of s squared equals the square root of the fraction ∑ open parenthesis x minus x ― close parenthesis squared over n minus 1")[$s = sqrt(s^(2)) = sqrt(frac(∑ ( x − limits(x)^(―) )^(2), n − 1))$]
]


    The #math.equation(block: false, alt: "n minus 1")[$n − 1$] on the bottom has to do with a concept called degrees of freedom. Basically, it makes the sample standard deviation a better approximation of the population standard deviation.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Population Variance]


      The #strong[population variance]formula:


      #math.equation(block: true, alt: "σ squared equals the fraction ∑ open parenthesis x minus μ close parenthesis squared over N")[$σ^(2) = frac(∑ ( x − μ )^(2), N)$]


      where #math.equation(block: false, alt: "σ")[$σ$] is the Greek letter sigma and #math.equation(block: false, alt: "σ squared")[$σ^(2)$] represents the population variance, #math.equation(block: false, alt: "μ")[$μ$] is the population mean, and N is the size of the population.
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Population Standard Deviation]


      The #strong[population standard deviation]formula:


      #math.equation(block: true, alt: "σ equals the square root of σ squared equals the square root of the fraction ∑ open parenthesis x minus μ close parenthesis squared over N")[$σ = sqrt(σ^(2)) = sqrt(frac(∑ ( x − μ )^(2), N))$]
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Note]


      The sum of the deviations should always be 0. If it isn’t, then it is because you rounded, you used the median instead of the mean, or you made an error. Try not to round too much in the calculations for standard deviation since each rounding causes a slight error
]


    #examplebox("Example 3")[Finding the Standard Deviation][
Suppose that a manager wants to test two new training programs. He randomly selects 5 people for each training type and measures the time it takes to complete a task after the training. The times for both trainings are in Example #math.equation(block: false, alt: "4")[$4$]. Which training method is better?


        #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Same mean, different spread]


          Training 1 is pre-loaded into L1 and Training 2 into L2 — switch the Data List and Calculate to reproduce both 1-Var Stats screens.


          - 1-Var Stats: s = 9.93 for Training 1, s = 3.35 for Training 2
]


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([Training 1], [56], [75], [48], [63], [59]),
  [Training 2], [60], [58], [66], [59], [58],
))


      
      #solutionbox[
It is important that you define what each variable is since there are two of them.


        Variable 1: #math.equation(block: false, alt: "X sub 1 equals")[$X_(1) =$] productivity from training 1


        Variable 2: #math.equation(block: false, alt: "X sub 2 equals")[$X_(2) =$] productivity from training 2


        To answer which training method better, first you need some descriptive statistics. Start with the mean for each sample.


        #math.equation(block: false, alt: "x ― sub 1 equals the fraction 56 plus 75 plus 48 plus 63 plus 59 over 5 equals 60.2")[$limits(x)^(―)_(1) = frac(56 + 75 + 48 + 63 + 59, 5) = 60.2$] minutes


        #math.equation(block: false, alt: "x ― sub 2 equals the fraction 60 plus 58 plus 66 plus 59 plus 58 over 5 equals 60.2")[$limits(x)^(―)_(2) = frac(60 + 58 + 66 + 59 + 58, 5) = 60.2$] minutes


        Since both means are the same values, you cannot answer the question about which is better. Now calculate the standard deviation for each sample.


        #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x sub 1")[$x_(1)$]], [#math.equation(block: false, alt: "x sub 1 minus x ― sub 1")[$x_(1) − limits(x)^(―)_(1)$]], [#math.equation(block: false, alt: "open parenthesis x sub 1 minus x ― sub 1 close parenthesis squared")[$attach(( x_(1) − limits(x)^(―)_(1) ), t: 2)$]]),
  [56], [-4.2], [17.64],
  [75], [14.8], [219.04],
  [48], [-12.2], [148.84],
  [63], [2.8], [7.84],
  [59], [-1.2], [1.44],
  [Total], [0], [394.8],
))


        #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x sub 2")[$x_(2)$]], [#math.equation(block: false, alt: "x sub 2 minus x ― sub 2")[$x_(2) − limits(x)^(―)_(2)$]], [#math.equation(block: false, alt: "open parenthesis x sub 2 minus x ― sub 2 close parenthesis squared")[$attach(( x_(2) − limits(x)^(―)_(2) ), t: 2)$]]),
  [60], [-0.2], [0.04],
  [58], [-2.2], [4.84],
  [66], [5.8], [33.64],
  [59], [-1.2], [1.44],
  [58], [-2.2], [4.84],
  [Total], [0], [44.8],
))


        The variance for each sample is:


        #math.equation(block: true, alt: "s 1 2 equals the fraction 394.8 over 5 minus 1 equals 98.7 minutes squared")[$s_(1)^(2) = frac(394.8, 5 − 1) = 98.7 attach(" minutes ", t: 2)$]


        #math.equation(block: true, alt: "s 2 2 equals the fraction 44.8 over 5 minus 1 equals 11.2 minutes squared")[$s_(2)^(2) = frac(44.8, 5 − 1) = 11.2 attach(" minutes ", t: 2)$]


        The standard deviations are:


        #math.equation(block: false, alt: "s sub 1 equals the square root of 98.7 approximately equals 9.93")[$s_(1) = sqrt(98.7) ≈ 9.93$] minutes


        #math.equation(block: false, alt: "s sub 2 equals the square root of 11.2 approximately equals 3.35")[$s_(2) = sqrt(11.2) ≈ 3.35$] minutes


        From the standard deviations, the second training seemed to be the better training since the data is less spread out. This means it is more consistent. It would be better for the managers in this case to have a training program that produces more consistent results so they know what to expect for the time it takes to complete the task.
]
]


    You can do the calculations for the descriptive statistics using the technology. The procedure for calculating the sample mean ( #math.equation(block: false, alt: "x ― close parenthesis")[$limits(x)^(―) \)$] and the sample standard deviation ( #math.equation(block: false, alt: "s sub x")[$s_(x)$]) for #math.equation(block: false, alt: "X sub 2")[$X_(2)$] in Example #math.equation(block: false, alt: "3")[$3$] on the TI-83/84 is in #emph[Figures 3.2.1] through #emph[3.2.4] (the procedure is the same for #math.equation(block: false, alt: "X sub 1")[$X_(1)$]). Note the calculator gives you the population standard deviation ( #math.equation(block: false, alt: "σ sub x")[$σ_(x)$] ) because it doesn’t know whether the data you input is a population or a sample. You need to decide which value you need to use, based on whether you have a population or sample. In almost all cases you have a sample and will be using #math.equation(block: false, alt: "s sub x")[$s_(x)$]. Also, the calculator uses the notation #math.equation(block: false, alt: "s sub x")[$s_(x)$] of instead of just #math.equation(block: false, alt: "s")[$s$]. It is just a way for it to denote the information. First you need to go into the STAT menu, and then Edit. This will allow you to type in your data (see #emph[Figure #math.equation(block: false, alt: "1")[$1$]]).


    #figure(figph[TI-84 calculator list editor screen showing values entered in L1 (56, 75, 48, 63, 59) and L2 (60, 58, 66, 59, 58), with the cursor on the sixth L2 entry.], alt: "TI-84 calculator list editor screen showing values entered in L1 (56, 75, 48, 63, 59) and L2 (60, 58, 66, 59, 58), with the cursor on the sixth L2 entry.", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: TI-83/84 Calculator Edit Setup])


    Once you have the data into the calculator, you then go back to the STAT menu, move over to CALC, and then choose 1-Var Stats (see#emph[Figure #math.equation(block: false, alt: "2")[$2$]]). The calculator will now put 1-Var Stats on the main screen. Now type in L2 (2nd button and 2) and then press ENTER. (Note if you have the newer operating system on the TI-84, then the procedure is slightly different.) The results from the calculator are in #emph[Figure #math.equation(block: false, alt: "4")[$4$]].


    #figure(figph[TI-83/84 calculator STAT menu with the CALC tab selected, listing 1-Var Stats, 2-Var Stats, Med-Med, LinReg(ax+b), QuadReg, CubicReg, and QuartReg (quartic regression).], alt: "TI-83/84 calculator STAT menu with the CALC tab selected, listing 1-Var Stats, 2-Var Stats, Med-Med, LinReg(ax+b), QuadReg, CubicReg, and QuartReg (quartic regression).", caption: [Figure #math.equation(block: false, alt: "2")[$2$]: TI-83/84 Calculator CALC Menu])


    #figure(figph[TI-84 calculator screen displaying the command header '1-Var Stats L2' with no results yet shown.], alt: "TI-84 calculator screen displaying the command header '1-Var Stats L2' with no results yet shown.", caption: [Figure #math.equation(block: false, alt: "3")[$3$]: TI-83/84 Calculator Input for Example #math.equation(block: false, alt: "3")[$3$] Variable #math.equation(block: false, alt: "X sub 2")[$X_(2)$]])


    #figure(figph[TI-84 calculator 1-Var Stats screen showing mean x̄=60.2, sums Σx=301 and Σx²=18165, sample standard deviation Sx=3.346640106, population standard deviation σx=2.993325909, and n=5.], alt: "TI-84 calculator 1-Var Stats screen showing mean x̄=60.2, sums Σx=301 and Σx²=18165, sample standard deviation Sx=3.346640106, population standard deviation σx=2.993325909, and n=5.", caption: [Figure #math.equation(block: false, alt: "4")[$4$]: TI-83/84 Calculator Results for Example #math.equation(block: false, alt: "3")[$3$] Variable #math.equation(block: false, alt: "X sub 2")[$X_(2)$]])


    The processes for finding the mean, median, range, standard deviation, and variance on R are as follows:


    variable\<-c(type in your data) #linebreak() 
To find the mean, use mean(variable) #linebreak() 
To find the median, use median(variable) #linebreak() 
To find the range, use range(variable). Then find maximum – minimum. #linebreak() 
To find the standard deviation, use sd(variable) #linebreak() 
To find the variance, use var(variable)


    For the second data set in Example #math.equation(block: false, alt: "3")[$3$], the commands and results would be


    productivity\_2\<-c(60, 58, 66, 59, 58) #linebreak() 
mean(productivity\_2) #linebreak() 
\[1\] 60.2 #linebreak() 
median(productivity\_2) #linebreak() 
\[1\] 59 #linebreak() 
range(productivity\_2) #linebreak() 
\[1\] 58 66 #linebreak() 
sd(productivity\_2) #linebreak() 
\[1\] 3.34664 #linebreak() 
var(productivity\_2) #linebreak() 
\[1\] 11.2


    productivity\_2 \<- c(60, 58, 66, 59, 58)
    mean(productivity\_2)
    median(productivity\_2)
    range(productivity\_2)
    sd(productivity\_2)
    var(productivity\_2)
    In general a “small” standard deviation means the data is close together (more consistent) and a “large” standard deviation means the data is spread out (less consistent). Sometimes you want consistent data and sometimes you don’t. As an example if you are making bolts, you want to lengths to be very consistent so you want a small standard deviation. If you are administering a test to see who can be a pilot, you want a large standard deviation so you can tell who are the good pilots and who are the bad ones.


    What do “small” and “large” mean? To a bicyclist whose average speed is 20 mph, s = 20 mph is huge. To an airplane whose average speed is 500 mph, s = 20 mph is nothing. The “size” of the variation depends on the size of the numbers in the problem and the mean. Another situation where you can determine whether a standard deviation is small or large is when you are comparing two different samples such as in example \#3.2.3. A sample with a smaller standard deviation is more consistent than a sample with a larger standard deviation.


    Many other books and authors stress that there is a computational formula for calculating the standard deviation. However, this formula doesn’t give you an idea of what standard deviation is and what you are doing. It is only good for doing the calculations quickly. It goes back to the days when standard deviations were calculated by hand, and the person needed a quick way to calculate the standard deviation. It is an archaic formula that this author is trying to eradicate it. It is not necessary anymore, since most calculators and computers will do the calculations for you with as much meaning as this formula gives. It is suggested that you never use it. If you want to understand what the standard deviation is doing, then you should use the definition formula. If you want an answer quickly, use a computer or calculator.


    
=== Use of Standard Deviation


    One of the uses of the standard deviation is to describe how a population is distributed by using Chebyshev’s Theorem. This theorem works for any distribution, whether it is skewed, symmetric, bimodal, or any other shape. It gives you an idea of how much data is a certain distance on either side of the mean.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Chebyshev's Theorem]


      For any set of data:


      - At least 75% of the data fall in the interval from #math.equation(block: false, alt: "μ minus 2 σ to μ plus 2 σ")[$μ − 2 σ " to " μ + 2 σ$].
- At least 88.9% of the data fall in the interval from #math.equation(block: false, alt: "μ minus 3 σ to μ plus 3 σ")[$μ − 3 σ " to " μ + 3 σ$].
- At least 93.8% of the data fall in the interval from #math.equation(block: false, alt: "μ minus 4 σ to μ plus 4 σ")[$μ − 4 σ " to " μ + 4 σ$].
]


    #examplebox("Example 4")[Using Chebyshev's Theorem][
The U.S. Weather Bureau has provided the information in Table 7 about the total annual number of reported strong to violent (F3+) tornados in the United States for the years 1954 to 2012. ("U.S. tornado climatology," 17).


        #figure(table(
  columns: 10, align: left, inset: 6pt,
  table.header([46], [47], [31], [41], [24], [56], [56], [23], [31], [59]),
  [39], [70], [73], [85], [33], [38], [45], [39], [35], [22],
  [51], [39], [51], [131], [37], [24], [57], [42], [28], [45],
  [98], [35], [54], [45], [30], [15], [35], [64], [21], [84],
  [40], [51], [44], [62], [65], [27], [34], [23], [32], [28],
  [41], [98], [82], [47], [62], [21], [31], [29], [32], [],
))


        + Use Chebyshev’s theorem to find an interval centered about the mean annual number of strong to violent (F3+) tornados in which you would expect at least 75% of the years to fall.
+ Use Chebyshev’s theorem to find an interval centered about the mean annual number of strong to violent (F3+) tornados in which you would expect at least 88.9% of the years to fall.


      
      #solutionbox[
a. Variable: #math.equation(block: false, alt: "x equals")[$x =$] number of strong or violent (F3+) tornadoes Chebyshev’s theorem says that at least 75% of the data will fall in the interval from #math.equation(block: false, alt: "μ minus 2 σ")[$μ − 2 σ$] to #math.equation(block: false, alt: "μ plus 2 σ")[$μ + 2 σ$].


        You do not have the population, so you need to estimate the population mean and standard deviation using the sample mean and standard deviation. You can find the sample mean and standard deviation using technology:


        #math.equation(block: true, alt: "x ― approximately equals 46.24 , s approximately equals 22.18")[$limits(x)^(―) ≈ 46.24 , s ≈ 22.18$]


        So,


        #math.equation(block: true, alt: "μ approximately equals 46.24 , σ approximately equals 22.18")[$μ ≈ 46.24 , σ ≈ 22.18$]


        #math.equation(block: true, alt: "μ minus 2 σ to μ plus 2 σ")[$μ − 2 σ " to " μ + 2 σ$]


        #math.equation(block: true, alt: "46.24 minus 2 open parenthesis 22.18 close parenthesis to 46.24 plus 2 open parenthesis 22.18 close parenthesis")[$46.24 − 2 ( 22.18 ) " to " 46.24 + 2 ( 22.18 )$]


        #math.equation(block: true, alt: "46.24 minus 44.36 to 46.24 plus 44.36")[$46.24 − 44.36 " to " 46.24 + 44.36$]


        #math.equation(block: true, alt: "1.88 to 90.60")[$1.88 " to " 90.60$]


        Since you can’t have fractional number of tornados, round to the nearest whole number.


        At least 75% of the years have between 2 and 91 strong to violent (F3+) tornados. (Actually, all but three years’ values fall in this interval, that means that #math.equation(block: false, alt: "the fraction 56 over 59 approximately equals 94.9")[$frac(56, 59) ≈ 94.9$] actually fall in the interval.)


        b. Variable: #math.equation(block: false, alt: "x equals")[$x =$] number of strong or violent (F3+) tornadoes Chebyshev’s theorem says that at least 88.9% of the data will fall in the interval from #math.equation(block: false, alt: "μ minus 3 σ")[$μ − 3 σ$] to #math.equation(block: false, alt: "μ plus 3 σ")[$μ + 3 σ$].


        #math.equation(block: true, alt: "μ minus 3 σ to μ plus 3 σ")[$μ − 3 σ " to " μ + 3 σ$]


        #math.equation(block: true, alt: "46.24 minus 3 open parenthesis 22.18 close parenthesis to 46.24 plus 3 open parenthesis 22.18 close parenthesis")[$46.24 − 3 ( 22.18 ) " to " 46.24 + 3 ( 22.18 )$]


        #math.equation(block: true, alt: "46.24 minus 66.54 to 46.24 plus 66.54")[$46.24 − 66.54 " to " 46.24 + 66.54$]


        #math.equation(block: true, alt: "minus 20.30 to 112.78")[$− 20.30 " to " 112.78$]


        Since you can’t have negative number of tornados, the lower limit is actually 0. Since you can’t have fractional number of tornados, round to the nearest whole number.


        At least 88.9% of the years have between 0 and 113 strong to violent (F3+) tornados.


        (Actually, all but one year falls in this interval, that means that #math.equation(block: false, alt: "the fraction 58 over 59 approximately equals 98.3")[$frac(58, 59) ≈ 98.3$] actually fall in the interval.)
]
]


    Chebyshev’s Theorem says that at least 75% of the data is within two standard deviations of the mean. That percentage is fairly high. There isn’t much data outside two standard deviations. A rule that can be followed is that if a data value is within two standard deviations, then that value is a common data value. If the data value is outside two standard deviations of the mean, either above or below, then the number is uncommon. It could even be called unusual. An easy calculation that you can do to figure it out is to find the difference between the data point and the mean, and then divide that answer by the standard deviation. As a formula this would be


    #math.equation(block: true, alt: "the fraction x minus μ over σ")[$frac(x − μ, σ)$].


    If you don’t know the population mean, #math.equation(block: false, alt: "μ")[$μ$], and the population standard deviation, #math.equation(block: false, alt: "σ")[$σ$], then use the sample mean, #math.equation(block: false, alt: "x ―")[$limits(x)^(―)$], and the sample standard deviation, #math.equation(block: false, alt: "s")[$s$], to estimate the population parameter values. However, realize that using the sample standard deviation may not actually be very accurate.


    #examplebox("Example 5")[determining if a value is unusual][
+ In 1974, there were 131 strong or violent (F3+) tornados in the United States. Is this value unusual? Why or why not?
+ In 1987, there were 15 strong or violent (F3+) tornados in the United States. Is this value unusual? Why or why not?


      
      #solutionbox[
a. Variable: #math.equation(block: false, alt: "x equals")[$x =$] number of strong or violent (F3+) tornadoes


        To answer this question, first find how many standard deviations 131 is from the mean. From Example #math.equation(block: false, alt: "4")[$4$], we know #math.equation(block: false, alt: "μ approximately equals 46.24")[$μ ≈ 46.24$] and #math.equation(block: false, alt: "σ approximately equals 22.18")[$σ ≈ 22.18$]. For #math.equation(block: false, alt: "x equals 131")[$x = 131$],


        #math.equation(block: true, alt: "the fraction x minus μ over σ equals the fraction 131 minus 46.24 over 22.18 approximately equals 3.82")[$frac(x − μ, σ) = frac(131 − 46.24, 22.18) ≈ 3.82$]


        Since this value is more than 2, then it is unusual to have 131 strong or violent (F3+) tornados in a year.


        b. Variable: #math.equation(block: false, alt: "x equals")[$x =$] number of strong or violent (F3+) tornadoes For this question the #math.equation(block: false, alt: "x equals 15")[$x = 15$],


        #math.equation(block: true, alt: "the fraction x minus μ over σ equals the fraction 15 minus 46.24 over 22.18 approximately equals minus 1.41")[$frac(x − μ, σ) = frac(15 − 46.24, 22.18) ≈ − 1.41$]


        Since this value is between -2 and 2, then it is not unusual to have only 15 strong or violent (F3+) tornados in a year.
]
]


    
=== Homework


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
+ Cholesterol levels were collected from patients two days after they had a heart attack (Ryan, Joiner & Ryan, Jr, 1985) and are in Table 8. Find the mean, median, range, variance, and standard deviation using technology.
    
    #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([270], [236], [210], [142], [280], [272], [160]),
  [220], [226], [242], [186], [266], [206], [318],
  [294], [282], [234], [224], [276], [282], [360],
  [310], [280], [278], [288], [288], [244], [236],
))
+ The lengths (in kilometers) of rivers on the South Island of New Zealand that flow to the Pacific Ocean are listed in Table 9 (Lee, 1994).
    

    a. Find the mean and median. #linebreak() 
    b. Find the range. #linebreak() 
    c. Find the variance and standard deviation.
    #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([River], [Length (km)], [River], [Length (km)]),
  [Clarence], [209], [Clutha], [322],
  [Conway], [48], [Taieri], [288],
  [Waiau], [169], [Shag], [72],
  [Hurunui], [138], [Kakanui], [64],
  [Waipara], [64], [Waitaki], [209],
  [Ashley], [97], [Waihao], [64],
  [Waimakariri], [161], [Pareora], [56],
  [Selwyn], [95], [Rangitata], [121],
  [Rakaia], [145], [Ophi], [80],
  [Ashburton], [90], [], [],
))
+ The lengths (in kilometers) of rivers on the South Island of New Zealand that flow to the Pacific Ocean are listed in Table 9 (Lee, 1994).
    

    Table #math.equation(block: false, alt: "10")[$10$]#emph[: Lengths of Rivers (km) Flowing to Tasman Sea] #linebreak() 
    a. Find the mean and median. #linebreak() 
    b. Find the range. #linebreak() 
    c. Find the variance and standard deviation.
    #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([River], [Length (km)], [River], [Length (km)]),
  [Hollyford], [76], [Waimea], [48],
  [Cascade], [64], [Motueka], [108],
  [Arawhata], [68], [Takaka], [72],
  [Haast], [64], [Aorere], [72],
  [Karangarua], [37], [Heaphy], [35],
  [Cook], [32], [Karamea], [80],
  [Waiho], [32], [Mokihinui], [56],
  [Whataroa], [51], [Buller], [177],
  [Wanganui], [56], [Grey], [121],
  [Waitaha], [40], [Taramakau], [80],
  [Hokitika], [64], [Arahura], [56],
))
+ Eyeglassmatic manufactures eyeglasses for their retailers. They test to see how many defective lenses they made the time period of January 1 to March 31. Table 11 gives the defect and the number of defects.
    
     #linebreak() 
    Table #math.equation(block: false, alt: "11")[$11$]#emph[: Number of Defective Lenses] #linebreak() 
    a. Find the mean and median. #linebreak() 
    b. Find the range. #linebreak() 
    c. Find the variance and standard deviation.#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Defect type], [Number of defects]),
  [Scratch], [5865],
  [Right shaped - small], [4613],
  [Flaked], [1992],
  [Wrong axis], [1838],
  [Chamfer wrong], [1596],
  [Crazing, cracks], [1546],
  [Wrong shape], [1485],
  [Wrong PD], [1398],
  [Spots and bubbles], [1371],
  [Wrong height], [1130],
  [Right shape - big], [1105],
  [Lost in lab], [976],
  [Spots/bubble - intern], [976],
))
+ Print-O-Matic printing company’s employees have salaries that are contained in Table 12. Find the mean, median, range, variance, and standard deviation using technology.
    
    #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Employee], [Salary (\$)], [Employee], [Salary (\$)]),
  [CEO], [272,500], [Administration], [66,346],
  [Driver], [58,456], [Sales], [109,739],
  [CD74], [100,702], [Designer], [90,090],
  [CD65], [57,380], [Platens], [69,573],
  [Embellisher], [73,877], [Polar], [75,526],
  [Folder], [65,270], [ITEK], [64,553],
  [GTO], [74,235], [Mgmt], [108,448],
  [Pre Press Manager], [108,448], [Handwork], [52,718],
  [Pre Press Manager/IT], [98,837], [Horizon], [76,029],
  [Pre Press/ Graphic Artist], [75,311], [], [],
))
+ Print-O-Matic printing company spends specific amounts on fixed costs every month. The costs of those fixed costs are in Table 13.
    

    
    a. Find the mean and median. #linebreak() 
    b. Find the range. #linebreak() 
    c. Find the variance and standard deviation.
    #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Monthly charges], [Monthly cost (\$)]),
  [Bank charges], [482],
  [Cleaning], [2208],
  [Computer expensive], [2471],
  [Lease payments], [2656],
  [Postage], [2117],
  [Uniforms], [2600],
))
+ Compare the two data sets in problems 2 and 3 using the mean and standard deviation. Discuss which mean is higher and which has a larger spread of the data.
+ Table 14 contains pulse rates collected from males, who are non-smokers but do drink alcohol ("Pulse rates before," 2013). The before pulse rate is before they exercised, and the after pulse rate was taken after the subject ran in place for one minute. Compare the two data sets using the mean and standard deviation. Discuss which mean is higher and which has a larger spread of the data.
    
    #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Pulse before], [Pulse after], [Pulse before], [Pulse after]),
  [76], [88], [59], [92],
  [56], [110], [60], [104],
  [64], [126], [65], [82],
  [50], [90], [76], [150],
  [49], [83], [145], [155],
  [68], [136], [84], [140],
  [68], [125], [78], [141],
  [88], [150], [85], [131],
  [80], [146], [78], [132],
  [78], [168], [], [],
))
+ Table 15 contains pulse rates collected from females, who are non-smokers but do drink alcohol ("Pulse rates before," 2013). The before pulse rate is before they exercised, and the after pulse rate was taken after the subject ran in place for one minute. Compare the two data sets using the mean and standard deviation. Discuss which mean is higher and which has a larger spread of the data.
    
    #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Pulse before], [Pulse after], [Pulse before], [Pulse after]),
  [96], [176], [92], [120],
  [82], [150], [70], [96],
  [86], [150], [75], [130],
  [72], [115], [70], [119],
  [78], [129], [70], [95],
  [90], [160], [68], [84],
  [88], [120], [47], [136],
  [71], [125], [64], [120],
  [66], [89], [70], [98],
  [76], [132], [74], [168],
  [70], [120], [85], [130],
))
+ To determine if Reiki is an effective method for treating pain, a pilot study was carried out where a certified second-degree Reiki therapist provided treatment on volunteers. Pain was measured using a visual analogue scale (VAS) immediately before and after the Reiki treatment (Olson & Hanson, 1997) and the data is in Table 16. Compare the two data sets using the mean and standard deviation. Discuss which mean is higher and which has a larger spread of the data.
    
    #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([VAS before], [VAS after], [VAS before], [VAS after]),
  [6], [3], [5], [1],
  [2], [1], [1], [0],
  [2], [0], [6], [4],
  [9], [1], [6], [1],
  [3], [0], [4], [4],
  [3], [2], [4], [1],
  [4], [1], [7], [6],
  [5], [2], [2], [1],
  [2], [2], [4], [3],
  [3], [0], [8], [8],
))
+ Table 17 contains data collected on the time it takes in seconds of each passage of play in a game of rugby. ("Time of passages," 2013)
    

    a. Using technology, find the mean and standard deviation. #linebreak() 
    b. Use Chebyshev’s theorem to find an interval centered about the mean times of each passage of play in the game of rugby in which you would expect at least 75% of the times to fall. #linebreak() 
    c. Use Chebyshev’s theorem to find an interval centered about the mean times of each passage of play in the game of rugby in which you would expect at least 88.9% of the times to fall.
    #figure(table(
  columns: 10, align: left, inset: 6pt,
  table.header([39.2], [2.7], [9.2], [14.6], [1.9], [17.8], [15.5], [53.8], [17.5], [27.5]),
  [4.8], [8.6], [22.1], [29.8], [10.4], [9.8], [27.7], [32.7], [32], [34.3],
  [29.1], [6.5], [2.8], [10.8], [9.2], [12.9], [7.1], [23.8], [7.6], [36.4],
  [35.6], [28.4], [37.2], [16.8], [21.2], [14.7], [44.5], [24.7], [36.2], [20.9],
  [19.9], [24.4], [7.9], [2.8], [2.7], [3.9], [14.1], [28.4], [45.5], [38],
  [18.5], [8.3], [56.2], [10.2], [5.5], [2.5], [46.8], [23.1], [9.2], [10.3],
  [10.2], [22], [28.5], [24], [17.3], [12.7], [15.5], [4], [5.6], [3.8],
  [21.6], [49.3], [52.4], [50.1], [30.5], [37.2], [15], [38.7], [3.1], [11],
  [10], [5], [48.8], [3.6], [12.6], [9.9], [58.6], [37.9], [19.4], [29.2],
  [12.3], [39.2], [22.2], [39.7], [6.4], [2.5], [34], [], [], [],
))
+ Yearly rainfall amounts (in millimeters) in Sydney, Australia, are in table \#3.2.18 ("Annual maximums of," 2013).
    

    a. Using technology, find the mean and standard deviation. #linebreak() 
    b. Use Chebyshev’s theorem to find an interval centered about the mean yearly rainfall amounts in Sydney, Australia, in which you would expect at least 75% of the amounts to fall. #linebreak() 
    c. Use Chebyshev’s theorem to find an interval centered about the mean yearly rainfall amounts in Sydney, Australia, in which you would expect at least 88.9% of the amounts to fall.
    #figure(table(
  columns: 8, align: left, inset: 6pt,
  table.header([146.8], [383], [90.9], [178.1], [267.5], [95.5], [156.5], [180]),
  [90.9], [139.7], [200.2], [171.7], [187.2], [184.9], [70.1], [58],
  [84.1], [55.6], [133.1], [271.8], [135.9], [71.9], [99.4], [110.6],
  [47.5], [97.8], [122.7], [58.4], [154.4], [173.7], [118.8], [88],
  [84.6], [171.5], [254.3], [185.9], [137.2], [138.9], [96.2], [85],
  [45.2], [74.7], [264.9], [113.8], [133.4], [68.1], [156.4], [],
))
+ The number of deaths attributed to UV radiation in African countries in the year 2002 is given in Table 19 ("UV radiation: Burden," 2013).
    

    a. Using technology, find the mean and standard deviation. #linebreak() 
    b. Use Chebyshev’s theorem to find an interval centered about the mean number of deaths from UV radiation in which you would expect at least 75% of the numbers to fall. #linebreak() 
    c. Use Chebyshev’s theorem to find an interval centered about the mean number of deaths from UV radiation in which you would expect at least 88.9% of the numbers to fall.
    #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([50], [84], [31], [338], [6], [504], [40], [7], [58]),
  [204], [15], [27], [39], [1], [45], [174], [98], [94],
  [199], [9], [27], [58], [356], [5], [45], [5], [94],
  [26], [171], [13], [57], [138], [39], [3], [171], [41],
  [1177], [102], [123], [433], [35], [40], [456], [125], [],
))
+ The time (in 1/50 seconds) between successive pulses along a nerve fiber ("Time between nerve," 2013) are given in Table 20.
    

    a. Using technology, find the mean and standard deviation. #linebreak() 
    b. Use Chebyshev’s theorem to find an interval centered about the mean time between successive pulses along a nerve fiber in which you would expect at least 75% of the times to fall. #linebreak() 
    c. Use Chebyshev’s theorem to find an interval centered about the mean time between successive pulses along a nerve fiber in which you would expect at least 88.9% of the times to fall.
    #figure(table(
  columns: 10, align: left, inset: 6pt,
  table.header([10.5], [1.5], [2.5], [5.5], [29.5], [3], [9], [27.5], [18.5], [4.5]),
  [7], [9.5], [1], [7], [4.5], [2.5], [7.5], [11.5], [7.5], [4],
  [12], [8], [3], [5.5], [7.5], [4.5], [1.5], [10.5], [1], [7],
  [12], [14.5], [8], [3.5], [3.5], [2], [1], [7.5], [6], [13],
  [7.5], [16.5], [3], [25.5], [5.5], [14], [18], [7], [27.5], [14],
))
+ Suppose a passage of play in a rugby game takes 75.1 seconds. Would it be unusual for this to happen? Use the mean and standard deviation that you calculated in problem 11.
+ Suppose Sydney, Australia received 300 mm of rainfall in a year. Would this be unusual? Use the mean and standard deviation that you calculated in problem 12.
+ Suppose in a given year there were 2257 deaths attributed to UV radiation in an African country. Is this value unusual? Use the mean and standard deviation that you calculated in problem 13.
+ Suppose it only takes 2 (1/50 seconds) for successive pulses along a nerve fiber. Is this value unusual? Use the mean and standard deviation that you calculated in problem 14.


      Answer


      1. mean = 253.93, median = 268, range = 218, variance = 2276.29, st dev = 47.71


      3. a. mean = 67.68 km, median = 64 km, b. range = 145 km, c. variance = 1107.9416 #math.equation(block: false, alt: "k m squared")[$attach(upright(k) upright(m), t: 2)$], st dev = 33.29 km


      5. mean = \$89,370.42, median = \$75,311, range = \$219,782, variance =2298639399, st dev = \$47,944.13


      7. See solutions


      9. #math.equation(block: false, alt: "x ― sub 1 approximately equals 75.45 , s sub 1 approximately equals 11.10 , x ― sub 2 approximately equals 125.55 , s sub 2 approximately equals 24.72")[$limits(x)^(―)_(1) ≈ 75.45 , s_(1) ≈ 11.10 , limits(x)^(―)_(2) ≈ 125.55 , s_(2) ≈ 24.72$]


      11. a. #math.equation(block: false, alt: "x ― approximately equals 21.24 s e c , s approximately equals 14.95 s e c")[$limits(x)^(―) ≈ 21.24 upright(s) upright(e) upright(c) , s ≈ 14.95 upright(s) upright(e) upright(c)$] b. #math.equation(block: false, alt: "open parenthesis minus 8.66 s e c , 51.14 s e c close parenthesis")[$( − 8.66 upright(s) upright(e) upright(c) , 51.14 upright(s) upright(e) upright(c) )$] c. #math.equation(block: false, alt: "open parenthesis minus 23.61 s e c , 66.09 s e c close parenthesis")[$( − 23.61 upright(s) upright(e) upright(c) , 66.09 upright(s) upright(e) upright(c) )$]


      13. a. #math.equation(block: false, alt: "x ― approximately equals 130.98 , s approximately equals 205.44")[$limits(x)^(―) ≈ 130.98 , s ≈ 205.44$] b. #math.equation(block: false, alt: "open parenthesis minus 279.90 , 541.86 close parenthesis")[$( − 279.90 , 541.86 )$] c. #math.equation(block: false, alt: "open parenthesis minus 485.34 , 747.3 close parenthesis")[$( − 485.34 , 747.3 )$]


      15. 3.61


      17. 10.35
]