#set document(title: "5.3 Mean and Standard Deviation of Binomial Distribution", author: "Kathryn Kozak") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 5.3#h(0.6em)Mean and Standard Deviation of Binomial Distribution If you list all possible values of #math.equation(block: false, alt: "x")[$x$] in a Binomial distribution, you get the #strong[Binomial Probability Distribution] (pdf). You can draw a histogram of the pdf and find the mean, variance, and standard deviation of it. For a general discrete probability distribution, you can find the mean, the variance, and the standard deviation for a pdf using the general formulas #math.equation(block: true, alt: "μ equals ∑ x P open parenthesis x close parenthesis , σ squared equals ∑ open parenthesis x minus μ close parenthesis squared P open parenthesis x close parenthesis , and σ equals the square root of ∑ open parenthesis x minus μ close parenthesis squared P open parenthesis x close parenthesis")[$μ = ∑ x P ( x ) , σ^(2) = ∑ ( x − μ )^(2) P ( x ) , " and " σ = sqrt(∑ ( x − μ )^(2) P ( x ))$] These formulas are useful, but if you know the type of distribution, like Binomial, then you can find the mean and standard deviation using easier formulas. They are derived from the general formulas. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] For a Binomial distribution, #math.equation(block: false, alt: "μ")[$μ$], the expected number of successes, #math.equation(block: false, alt: "σ squared")[$σ^(2)$], the variance, and #math.equation(block: false, alt: "σ")[$σ$], the standard deviation for the number of success are given by the formulas: #math.equation(block: true, alt: "μ equals n p σ squared equals n p q σ equals the square root of n p q")[$μ = n p #h(1em) σ^(2) = n p q #h(1em) σ = sqrt(n p q)$] Where p is the probability of success and #emph[q]= 1 - #emph[p]. ] #examplebox("Example 1")[Finding the Probability Distribution, Mean, Variance, and Standard Deviation of a Binomial Distribution][ When looking at a person’s eye color, it turns out that 1% of people in the world has green eyes ("What percentage of," 2013). Consider a group of 20 people. + State the random variable. + Write the probability distribution. + Draw a histogram. + Find the mean. + Find the variance. + Find the standard deviation. #solutionbox[ a. #emph[x] = number of people who have green eyes b. In this case you need to write each value of #emph[x] and its corresponding probability. It is easiest to do this by using the binompdf command, but don’t put in the #emph[r] value. You may want to set your calculator to only three decimal places, so it is easier to see the values and you don’t need much more precision than that. The command would look like #math.equation(block: false, alt: "binompdf open parenthesis 20 , .01 close parenthesis")[$"binompdf" ( 20 , ".01" )$]. This produces the information in Example #math.equation(block: false, alt: "1")[$1$]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#emph[x]], [#emph[P] (x=r)]), [0], [0.818], [1], [0.165], [2], [0.016], [3], [0.001], [4], [0.000], [5], [0.000], [6], [0.000], [7], [0.000], [8], [0.000], [9], [0.000], [10], [0.000], [#math.equation(block: false, alt: "⋮")[$⋮$]], [#math.equation(block: false, alt: "⋮")[$⋮$]], [20], [0.000], )) Notice that after x = 4, the probability values are all 0.000. This just means they are really small numbers. c. You can draw the histogram on the TI-83/84 or other technology. The graph would look like in #emph[Figure #math.equation(block: false, alt: "1")[$1$]]. #figure(figph[Bar chart of a binomial probability distribution with probability 0.818 at x=0, 0.165 at x=1, 0.016 at x=2, 0.001 at x=3, and zero thereafter through x=20.], alt: "Bar chart of a binomial probability distribution with probability 0.818 at x=0, 0.165 at x=1, 0.016 at x=2, 0.001 at x=3, and zero thereafter through x=20.", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: Histogram Created on TI-83/84]) This graph is very skewed to the right. d. Since this is a binomial, then you can use the formula #math.equation(block: false, alt: "μ equals n p")[$μ = n p$]. So #math.equation(block: false, alt: "μ equals 20 open parenthesis 0.01 close parenthesis equals 0.2")[$μ = 20 ( 0.01 ) = 0.2$] people. You expect on average that out of 20 people, less than 1 would have green eyes. e. Since this is a binomial, then you can use the formula #math.equation(block: false, alt: "σ squared equals n p q")[$σ^(2) = n p q$]. #math.equation(block: true, alt: "q equals 1 minus 0.01 equals 0.99")[$q = 1 − 0.01 = 0.99$] #math.equation(block: true, alt: "σ squared equals 20 open parenthesis 0.01 close parenthesis open parenthesis 0.99 close parenthesis equals 0.198 people squared")[$σ^(2) = 20 ( 0.01 ) ( 0.99 ) = 0.198 attach(" people ", t: 2)$] f. Once you have the variance, you just take the square root of the variance to find the standard deviation. #math.equation(block: true, alt: "σ equals the square root of 0.198 approximately equals 0.445")[$σ = sqrt(0.198) ≈ 0.445$] ] ] === Homework #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Suppose a random variable, #emph[x], arises from a binomial experiment. Suppose #emph[n] = 6, and #emph[p] = 0.13. + Write the probability distribution. + Draw a histogram. + Describe the shape of the histogram. + Find the mean. + Find the variance. + Find the standard deviation. + Suppose a random variable, #emph[x], arises from a binomial experiment. Suppose #emph[n] = 10, and #emph[p] = 0.81. + Write the probability distribution. + Draw a histogram. + Describe the shape of the histogram. + Find the mean. + Find the variance. + Find the standard deviation. + Suppose a random variable, #emph[x], arises from a binomial experiment. Suppose #emph[n] = 7, and #emph[p] = 0.50. + Write the probability distribution. + Draw a histogram. + Describe the shape of the histogram. + Find the mean. + Find the variance. + Find the standard deviation. + Approximately 10% of all people are left-handed. Consider a grouping of fifteen people. + State the random variable. + Write the probability distribution. + Draw a histogram. + Describe the shape of the histogram. + Find the mean. + Find the variance. + Find the standard deviation. + According to an article in the American Heart Association’s publication Circulation, 24% of patients who had been hospitalized for an acute myocardial infarction did not fill their cardiac medication by the seventh day of being discharged (Ho, Bryson & Rumsfeld, 2009). Suppose there are twelve people who have been hospitalized for an acute myocardial infarction. + State the random variable. + Write the probability distribution. + Draw a histogram. + Describe the shape of the histogram. + Find the mean. + Find the variance. + Find the standard deviation. + Eyeglassomatic manufactures eyeglasses for different retailers. In March 2010, they tested to see how many defective lenses they made, and there were 16.9% defective lenses due to scratches. Suppose Eyeglassomatic examined twenty eyeglasses. + State the random variable. + Write the probability distribution. + Draw a histogram. + Describe the shape of the histogram. + Find the mean. + Find the variance. + Find the standard deviation. + The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s. + State the random variable. + Find the mean. + Find the variance. + Find the standard deviation. Answer 1. a. See solutions, b. See solutions, c. Skewed right, d. 0.78, e. 0.6786, f. 0.8238 3. a. See solutions, b. See solutions, c. Symmetric, d. 3.5, e. 1.75, f. 1.3229 5. a. See solutions, b. See solutions, c. See solutions, d. Skewed right, e. 2.88, f. 2.1888, g. 1.479 7. a. See solutions, b. 7.28, c. 6.2608, d. 2.502 ] Data Sources: #emph[11 little-known facts about left-handers]. (2013, October 21). Retrieved from www.huffingtonpost.com/2012/1...n\_2005864.html #emph[CDC-data and statistics, autism spectrum disorders - ncbdd]. (2013, October 21). Retrieved from #link("http://www.cdc.gov/ncbddd/autism/data.html")[http://www.cdc.gov/ncbddd/autism/data.html] Ho, P. M., Bryson, C. L., & Rumsfeld, J. S. (2009). Medication adherence. #emph[Circulation], #emph[119] (23), 3028-3035. Retrieved from #link("http://circ.ahajournals.org/content/119/23/3028")[http://circ.ahajournals.org/content/119/23/3028] #emph[Households by age of householder and size of household: 1990 to 2010]. (2013, October 19). Retrieved from www.census.gov/compendia/stat...es/12s0062.pdf Madison, J. (2013, October 15). #emph[M&M's color distribution analysis.]Retrieved from #link("http://joshmadison.com/2007/12/02/mms-color-distribution-analysis/")[http://joshmadison.com/2007/12/02/mm...tion-analysis/] #emph[What percentage of people have green eyes?.] (2013, October 21). 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