#set document(title: "6.3 Finding Probabilities for the Normal 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")) == 6.3#h(0.6em)Finding Probabilities for the Normal Distribution The Empirical Rule is just an approximation and only works for certain values. What if you want to find the probability for #emph[x] values that are not integer multiples of the standard deviation? The probability is the area under the curve. To find areas under the curve, you need calculus. Before technology, you needed to convert every #emph[x] value to a standardized number, called the #emph[z]-score or #emph[z]-value or simply just #emph[z]. The #emph[z]-score is a measure of how many standard deviations an #emph[x] value is from the mean. To convert from a normally distributed #emph[x] value to a #emph[z]-score, you use the following formula. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[z-score] #math.equation(block: true, alt: "z equals the fraction x minus μ over σ")[$z = frac(x − μ, σ)$] where #math.equation(block: false, alt: "μ")[$μ$]= mean of the population of the x value and #math.equation(block: false, alt: "σ")[$σ$]= standard deviation for the population of the x value ] The z-score is normally distributed, with a mean of 0 and a standard deviation of 1. It is known as the standard normal curve. Once you have the z-score, you can look up the z-score in the standard normal distribution table. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[standard normal distribution] The#strong[standard normal distribution], z, has a mean of #math.equation(block: false, alt: "μ equals 0")[$μ = 0$] and a standard deviation of #math.equation(block: false, alt: "σ equals 1")[$σ = 1$]. #figure(figph[Normal curve for the standard normal distribution with mean z = 0 and one standard deviation extending from z = 0 to z = 1.], alt: "Normal curve for the standard normal distribution with mean z = 0 and one standard deviation extending from z = 0 to z = 1.", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: Standard Normal Curve]) ] Luckily, these days technology can find probabilities for you without converting to the z-score and looking the probabilities up in a table. There are many programs available that will calculate the probability for a normal curve including Excel and the TI-83/84. There are also online sites available. The following examples show how to do the calculation on the TI-83/84 and with R. The command on the TI-83/84 is in the DISTR menu and is normalcdf(. You then type in the lower limit, upper limit, mean, standard deviation in that order and including the commas. The command on R to find the area to the left is pnorm(z-value or x-value, mean, standard deviation). #examplebox("Example 1")[general normal distribution][ The length of a human pregnancy is normally distributed with a mean of 272 days with a standard deviation of 9 days (Bhat & Kushtagi, 2006). #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Every normalcdf screen in this example, live] Each link opens the Normal panel pre-loaded on N(272, 9) with the area already computed and drawn — the picture the example asks you to sketch, plus the calculator's number. - P(x \> 280) = 0.1870 — normalcdf(280, 1E99, 272, 9) - P(x \< 250) = 0.0073 — normalcdf(−1E99, 250, 272, 9) - P(265 \< x \< 280) = 0.5946 — normalcdf(265, 280, 272, 9) - invNorm(0.1, 272, 9) = 260.47 days ] + State the random variable. + Find the probability of a pregnancy lasting more than 280 days. + Find the probability of a pregnancy lasting less than 250 days. + Find the probability that a pregnancy lasts between 265 and 280 days. + Find the length of pregnancy that 10% of all pregnancies last less than. + Suppose you meet a woman who says that she was pregnant for less than 250 days. Would this be unusual and what might you think? #solutionbox[ a. #emph[x] = length of a human pregnancy b. First translate the statement into a mathematical statement. #emph[P] (x\>280) Now, draw a picture. Remember the center of this normal curve is 272. #figure(figph[A normal curve centered at 272 with the right-tail area shaded for values greater than 280.], alt: "A normal curve centered at 272 with the right-tail area shaded for values greater than 280.", caption: [Figure for Example #math.equation(block: false, alt: "1")[$1$]b]) To find the probability on the TI-83/84, looking at the picture you realize the lower limit is 280. The upper limit is infinity. The calculator doesn’t have infinity on it, so you need to put in a really big number. Some people like to put in 1000, but if you are working with numbers that are bigger than 1000, then you would have to remember to change the upper limit. The safest number to use is #math.equation(block: false, alt: "1 times 10 to the power 99")[$1 × 10^(99)$], which you put in the calculator as 1E99 (where E is the EE button on the calculator). The command looks like: #math.equation(block: true, alt: "normalcdf open parenthesis 280 , 1 E 99 , 272 , 9 close parenthesis")[$"normalcdf" ( 280 , 1 E 99 , 272 , 9 )$] #figure(figph[TI-84 calculator screen showing the normalcdf(280,1E99,272,9) calculation with result 0.1870313608.], alt: "TI-84 calculator screen showing the normalcdf(280,1E99,272,9) calculation with result 0.1870313608.", caption: [Figure #math.equation(block: false, alt: "3")[$3$]: TI-83/84 Output for Example #math.equation(block: false, alt: "1")[$1$]b]) To find the probability on R, R always gives the probability to the left of the value. The total area under the curve is 1, so if you want the area to the right, then you find the area to the left and subtract from 1. The command looks like: #math.equation(block: true, alt: "1 minus pnom open parenthesis 280 , 272 , 9 close parenthesis")[$1 − " pnom " ( 280 , 272 , 9 )$] Thus, #math.equation(block: false, alt: "P open parenthesis x greater than 280 close parenthesis approximately equals 0.187")[$P ( x > 280 ) ≈ 0.187$] Thus 18.7% of all pregnancies last more than 280 days. This is not unusual since the probability is greater than 5%. c. First translate the statement into a mathematical statement. #emph[P] (x\<250) Now, draw a picture. Remember the center of this normal curve is 272. #figure(figph[A normal distribution curve centered at 272 with an approximate standard deviation of 9.], alt: "A normal distribution curve centered at 272 with an approximate standard deviation of 9.", caption: [Figure for Example #math.equation(block: false, alt: "1")[$1$]c]) To find the probability on the TI-83/84, looking at the picture, though it is hard to see in this case, the lower limit is negative infinity. Again, the calculator doesn’t have this on it, put in a really small number, such as #math.equation(block: false, alt: "minus 1 times 10 to the power 99 equals minus 1 E 99")[$− 1 × 10^(99) = − 1 E 99$] on the calculator. #figure(figph[TI-83/84 calculator screen showing the normalcdf command with lower bound −1E99, upper bound 250, mean 272, standard deviation 9, and result 0.0072537738.], alt: "TI-83/84 calculator screen showing the normalcdf command with lower bound −1E99, upper bound 250, mean 272, standard deviation 9, and result 0.0072537738.", caption: [Figure #math.equation(block: false, alt: "5")[$5$]: TI-83/84 Output for Example #math.equation(block: false, alt: "1")[$1$]c]) #math.equation(block: true, alt: "P open parenthesis x less than 250 close parenthesis equals normalcdf open parenthesis minus 1 E 99 , 250 , 272 , 9 close parenthesis equals 0.0073")[$P ( x < 250 ) = " normalcdf " ( − 1 E 99 , 250 , 272 , 9 ) = 0.0073$] To find the probability on R, R always gives the probability to the left of the value. Looking at the figure, you can see the area you want is to the left. The command looks like: #math.equation(block: true, alt: "P open parenthesis x less than 250 close parenthesis equals pnorm open parenthesis 250 , 272 , 9 close parenthesis equals 0.0073")[$P ( x < 250 ) = " pnorm " ( 250 , 272 , 9 ) = 0.0073$] pnorm(250, 272, 9) Thus 0.73% of all pregnancies last less than 250 days. This is unusual since the probability is less than 5%. d. First translate the statement into a mathematical statement. #emph[P] (265\700) Now, draw a picture. Remember the center of this normal curve is 514. #figure(figph[A normal curve centered at 514 with the right-tail area shaded for values greater than 700.], alt: "A normal curve centered at 514 with the right-tail area shaded for values greater than 700.", caption: [Figure for Example #math.equation(block: false, alt: "2")[$2$]b]) On TI-83/84: #math.equation(block: false, alt: "P open parenthesis x greater than 700 close parenthesis equals normalcdf open parenthesis 700 , 1 E 99 , 514 , 117 close parenthesis approximately equals 0.056")[$P ( x > 700 ) = " normalcdf " ( 700 , 1 E 99 , 514 , 117 ) ≈ 0.056$] On R: #math.equation(block: false, alt: "P open parenthesis x greater than 700 close parenthesis equals 1 minus pnorm open parenthesis 700 , 514 , 117 close parenthesis approximately equals 0.056")[$P ( x > 700 ) = 1 − " pnorm " ( 700 , 514 , 117 ) ≈ 0.056$] pnorm(700, 514, 117) There is a 5.6% chance that a person scored above a 700 on the mathematics SAT test. This is not unusual. c. First translate the statement into a mathematical statement. #emph[P] (x\<400) Now, draw a picture. Remember the center of this normal curve is 514. #figure(figph[A normal curve centered at 514 with the area to the left of 414 shaded in blue.], alt: "A normal curve centered at 514 with the area to the left of 414 shaded in blue.", caption: [Figure for Example #math.equation(block: false, alt: "2")[$2$]c]) On TI-83/84: #math.equation(block: false, alt: "P open parenthesis x less than 400 close parenthesis equals normalcdf open parenthesis minus 1 E 99 , 400 , 514 , 117 close parenthesis approximately equals 0.165")[$P ( x < 400 ) = " normalcdf " ( − 1 E 99 , 400 , 514 , 117 ) ≈ 0.165$] On R: #math.equation(block: false, alt: "P open parenthesis x less than 400 close parenthesis equals pnorm open parenthesis 400 , 514 , 117 close parenthesis approximately equals 0.165")[$P ( x < 400 ) = op("pnorm") ( 400 , 514 , 117 ) ≈ 0.165$] pnorm(400, 514, 117) So, there is a 16.5% chance that a person scores less than a 400 on the mathematics part of the SAT. d. First translate the statement into a mathematical statement. #emph[P] (500\0.67) + #emph[P] (0\ 0.67 ) = 0.2514$], c. #math.equation(block: false, alt: "P open parenthesis 0 less than z less than 2.11 close parenthesis equals 0.4826")[$P ( 0 < z < 2.11 ) = 0.4826$], d. #math.equation(block: false, alt: "P open parenthesis minus 2.78 less than z less than 1.97 close parenthesis equals 0.9729")[$P ( − 2.78 < z < 1.97 ) = 0.9729$] 3. a. -0.6667, b. -2.6667, c. -2, d. 6.6667 5. a. See solutions, b. #math.equation(block: false, alt: "P open parenthesis x less than 52 c m close parenthesis equals 0.7128")[$P ( x < 52 upright(c) upright(m) ) = 0.7128$], c. #math.equation(block: false, alt: "P open parenthesis x greater than 74 c m close parenthesis equals 5.852 times 10 to the power minus 11")[$P ( x > 74 upright(c) upright(m) ) = 5.852 × 10^(− 11)$], d. #math.equation(block: false, alt: "P open parenthesis 40.5 c m less than x less than 57.5 c m close parenthesis equals 0.9729")[$P ( 40.5 upright(c) upright(m) < x < 57.5 upright(c) upright(m) ) = 0.9729$], e. See solutions, f. 53.8 cm 7. a. See solutions, b. #math.equation(block: false, alt: "P open parenthesis x greater than 110 g close parenthesis equals 0.0551")[$P ( x > 110 g ) = 0.0551$] c. #math.equation(block: false, alt: "P open parenthesis x less than 93 g close parenthesis equals 0.0097")[$P ( x < 93 g ) = 0.0097$], d. #math.equation(block: false, alt: "P open parenthesis x less than 65 g close parenthesis approximately equals 0")[$P ( x < 65 g ) ≈ 0$] or #math.equation(block: false, alt: "5.57 times 10 to the power minus 19")[$5.57 × 10^(− 19)$], e. See solutions, f. 110.2 g 9. a. See solutions, b. #math.equation(block: false, alt: "P open parenthesis x greater than $ 80 , 000 close parenthesis equals 0.1168")[$P ( x > \$ 80 , 000 ) = 0.1168$], c. #math.equation(block: false, alt: "P open parenthesis x greater than $ 80 , 000 close parenthesis equals 0.2283")[$P ( x > \$ 80 , 000 ) = 0.2283$], d. #math.equation(block: false, alt: "P open parenthesis $ 55 , 000 less than x less than $ 72 , 000 close parenthesis equals 0.5519")[$P ( \$ 55 , 000 < x < \$ 72 , 000 ) = 0.5519$], e. See solutions, f. \$73,112 ]