Converting from a N(, 2) Distributionto a N(0, 1) DistributionIf X ~ N(, 2), then what is Pr (a < X < b) for any a, b? Suppose serum total cholesterol is N(200, 400), what is the probability that a randomly selected person from this population will have mild hypercholesterolemia? That is, what is Pr (220 < X < 260)?We only have tables for N(0, 1) distribution. We must convert the probability statement for N(, 2) to a probability statement for N(0, 1). Let’s create a new random variable Z such thatσμXZZ ~ N(0, 1).)aZPr()bZPr(orσμaΦσμbΦσμbZσμaPrb)X(aPrTherefore,This is the cumulative distribution function for a N(0, 1) as given in column A of Table 3 in the Appendix.  = 200, 2 = 400 implies  = 20, a = 220, b = 260.003202002600012020022021.σμbZ.σμaZWe have converted the probability statement about N(200, 400) to one about N(0, 1). We can evaluate by using the standard normal tables. 1574.08413.09987.0)00.1()00.3(or)00.1ZPr()00.3ZPr()00.3Z00.1(Pr20200260Z20200220Pr)260X220(PrExample: Suppose we have a population of sucrose concentrations that is normally distributed with  = 65 mg/100 ml and  = 25 mg/100 ml. What proportion of the population is greater than 85 mg/100 ml? X ~ N(65, 625). Find Pr(X > 85).  80025658585 .ZZ)(X  PrPrPrUsing Table 3, column B, Pr (Z > 0.80) = 0.2119. Therefore, about 21.19% of the population has a sucrose concentration greater than 85 mg/100 ml. What proportion of the population is less than 45 mg/100 ml? 80025654545 .ZZ)(X  PrPrPrUsing Table 3, column A, Pr (Z < -0.80) = 1- Pr (Z < 0.80) = 1 - 0.7881 = 0.2119 orUsing Table 3, column B, Pr (Z < -0.80) = Pr (Z > 0.80) = 0.2119What proportion of the population is between 45 mg/100 ml and 85 mg/100 ml? 8008002565852565458545.Z.Z)X(PrPrPrUsing Table 3, column D, Pr (-0.80 < Z < 0.80) = .5763 orUsing Table 3, column A, Pr (-0.80 < Z < 0.80) =Pr (Z < 0.80) – Pr (Z < -0.80) = (0.80) - (-0.80) = 0.7881 – (1-0.7881) = 0.7881 – 0.2119 = 0.5762Example: Suppose the height (cm) of a populationis N(150, 400). In a sample of 5000 people, what percent would be taller than 160 cm?).(Z).(ZZ)(X500150020150160160PrPrPrPrUsing Table 3, column A, 1 – Pr(Z < 0.50) = 1- 0.6915 = 0.3085.The number taller than 160 cm would be 5000 * 0.3085 = 1542.5 or about 1542 people.NORMAL APPROXIMATION TO THEBINOMIALBinomial: Probability of k successes in n independent trials where p is the same in each trial. When n is large, there are no tables and using the formula is cumbersome. Therefore, we can use the Normal Distribution as an approximation to the Binomial Distribution. The Normal Distribution is a good approximation if n is large and p not too near 0 or 1. The binomial then will be nearly symmetric. Binomial: mean = np, variance = npq. Let’s create a normal distribution that is N(np, npq).We want Pr (a  X  b) where X is binomially distributed with parameters n and p. The first choice for an approximation is to examine the N(np, npq) normal distribution by getting the area under this curve from a to b. Continuity Correction: Better to get area under the normal curve from a – ½ to b + ½.Note: Pr (X = a) has a meaning for the binomial distribution but not for the normal distribution. If a = b, Pr (X = a) can be approximated by the area under the normal distribution from a – ½ to a + ½. For Pr (X = 0), the approximation under the normal distribution is the area under the normal distribution to the left of ½. For Pr (X = n), the approximation under the normal distribution is the area to the right of n – ½. Example: Suppose we have a binomial distribution with n = 20 and p = 0.30. Mean = np = 20 * 0.30 = 6Var = npq = 20 * 0.30 * 0.70 = 4.2049224 ..SD Suppose in 20 trials we have exactly 5 successes. Find Pr (X = 5) first using the Binomial Distribution and then using the Normal Distribution approximation.Binomial: Table 1, n = 20, p = 0.30, k = 5. Pr (X = 5) = 0.1789Normal: We want the area under the normal distribution curve from 4.5 to 5.5 or 1725.02327.04052.0Bcolumn3,Tableusing)73.0(ZPr)24.0(ZPr)73.0(ZPr)24.0(ZPr)73.0()24.0()24.0Z73.0(Pr049.265.5Z049.265.4Pr)5.5X5.4(PrNotice that the exact probability from the Binomial tables was 0.1789. This is a fairly good approximation even though n is small. The normal distribution, N(np, npq) is a good approximation to the binomial with parameters n and p when n*p*q  5. For the example, n*p*q = 20 * 0.30 * 0.70 = 4.2.Example: The rate of operative complications in acertain complex vascular reconstructive procedure is 20%. In a series of 50 such procedures, what is the probability that there will be at most 5 patients with operative complication?Pr(at most 5 complications) = Pr (X  5). p = 0.20, n = 50. We do not have tables. We could use formula for binomial for X = 0, X = 1, … X = 5 and then sum. Normal approximation: np = 10, npq = 8, SD =