#set document(title: "7.6 Normal Approximation to the Binomial", author: "OpenStax") #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")) == 7.6#h(0.6em)Normal Approximation to the Binomial #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Prerequisites] Binomial Distribution, History of the Normal Distribution, Areas of Normal Distributions #linebreak() #linebreak() ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Learning Objectives] + State the relationship between the normal distribution and the binomial distribution + Use the normal distribution to approximate the binomial distribution + State when the approximation is adequate ] In the section on the history of the normal distribution, we saw that the normal distribution can be used to approximate the binomial distribution. This section shows how to compute these approximations. Let's begin with an example. Assume you have a fair coin and wish to know the probability that you would get 8 heads out of 10 flips. The binomial distribution has a mean of μ = Nπ = (10)(0.5) = 5 and a variance of σ#super[2] = Nπ(1-π) = (10)(0.5)(0.5) = 2.5. The standard deviation is therefore 1.5811. A total of 8 heads is (8 - 5)/1.5811 = 1.897 standard deviations above the mean of the distribution. The question then is, "What is the probability of getting a value exactly 1.897 standard deviations above the mean?" You may be surprised to learn that the answer is 0: The probability of any one specific point is 0. The problem is that the binomial distribution is a discrete probability distribution, whereas the normal distribution is a continuous distribution. The solution is to round off and consider any value from 7.5 to 8.5 to represent an outcome of 8 heads. Using this approach, we figure out the area under a normal curve from 7.5 to 8.5. The area in green in Figure 1 is an approximation of the probability of obtaining 8 heads. #figure(figph[The binomial distribution for 10 coin flips drawn as blue spikes at 0 through 10 with a normal curve over them, and the strip of area under the curve from 7.5 to 8.5 shaded green around a red spike at 8. The green strip is the continuity-corrected normal approximation to the probability of exactly 8 heads.], alt: "The binomial distribution for 10 coin flips drawn as blue spikes at 0 through 10 with a normal curve over them, and the strip of area under the curve from 7.5 to 8.5 shaded green around a red spike at 8. The green strip is the continuity-corrected normal approximation to the probability of exactly 8 heads.", caption: [Figure 1. Approximating the probability of 8 heads with the normal distribution.]) The solution is therefore to compute this area. First we compute the area below 8.5 and then subtract the area below 7.5. The results of using the normal area calculator to find the area below 8.5 are shown in Figure 2. The results for 7.5 are shown in Figure 3. #figure(figph[Screenshot of the normal area calculator with Mean 5 and Sd 1.5811 — the mean and standard deviation of the binomial for 10 flips — and Below set to 8.5. Nearly the whole curve is shaded blue and the result reads Shaded area: 0.986574.], alt: "Screenshot of the normal area calculator with Mean 5 and Sd 1.5811 — the mean and standard deviation of the binomial for 10 flips — and Below set to 8.5. Nearly the whole curve is shaded blue and the result reads Shaded area: 0.986574.", caption: [Figure 2. Area below 8.5.]) #figure(figph[Screenshot of the same calculator with Mean 5 and Sd 1.5811 and Below set to 7.5. Slightly less of the curve is shaded and the result reads Shaded area: 0.943081. Subtracting this from the 0.986574 of the previous figure gives the strip from 7.5 to 8.5, about 0.044.], alt: "Screenshot of the same calculator with Mean 5 and Sd 1.5811 and Below set to 7.5. Slightly less of the curve is shaded and the result reads Shaded area: 0.943081. Subtracting this from the 0.986574 of the previous figure gives the strip from 7.5 to 8.5, about 0.044.", caption: [Figure 3. Area below 7.5.]) The difference between the areas is 0.044, which is the approximation of the binomial probability. For these parameters, the approximation is very accurate. The demonstration in the next section allows you to explore its accuracy with different parameters. If you did not have the normal area calculator, you could find the solution using a table of the standard normal distribution (a Z table) as follows: + Find a Z score for 8.5 using the formula Z = (8.5 - 5)/1.5811 = 2.21. + Find the area below a Z of 2.21 = 0.987. + Find a Z score for 7.5 using the formula Z = (7.5 - 5)/1.5811 = 1.58. + Find the area below a Z of 1.58 = 0.943. + Subtract the value in step 4 from the value in step 2 to get 0.044. The same logic applies when calculating the probability of a range of outcomes. For example, to calculate the probability of 8 to 10 flips, calculate the area from 7.5 to 10.5. The accuracy of the approximation depends on the values of N and π. A rule of thumb is that the approximation is good if both Nπ and N(1-π) are both greater than 10.