( ( What we will do today ly Rom Stard ( David Meredith Department of Mathematics San Francisco State University October 6, 2009 ly Rom Stard 1 ly Rom 2 3 Stard 4 ( ( Rom ly Rom Stard A variable is a characteristic of a population A quantitative variable is a numerical characteristic A normally distributed variable is a quantitative variable that fits a certain pattern Let µ be the average σ the stard deviation of the variable 68% of the data are in the interval (µ σ, µ + σ) 95% of the data are in the interval (µ 2σ, µ + 2σ) 99.7% of the data are in the interval (µ 3σ, µ + 3σ) We are going to learn to answer more detailed questions What percent of the data are greater than any number? What percent of the data are less than any number? 90% of the data are below what value ly Rom Stard We really don t care what the data is: heights, weights, wages,... All we need to know is that the numbers are normally distributed with some given mean µ stard deviation σ A normal rom variable X represents a set of numbers with a normal distribution. Usually we will say something like: let X be a normally distributed rom variable with mean µ = 3 stard deviation σ = 0.75.
( How to use the symbol X ( Finding P(X < a) ly Rom Stard We never say: What does X equal X sts for an unknown, romly chosen element of the distribution We say things like: What proportion of the distribution is more than 4: P(X > 4) What proportion of the distribution is between 2 3.5: P(2 < X < 3.5). 10% of the distribution is above what number: P(X >?) = 0.1 or P(X > x) = 0.1; find x. ly Rom Stard We need a practical way to find P(X < a) You have several tools you can use: R Excel Your calculator table We will use technology first, then table. ( Using R Excel to find P(X < a) ( How to find P(X > a) ly Rom Stard Suppose X is a normal rom variable with mean µ = 3 stard deviation σ = 0.75. To find P(X < a) with R, enter: pnorm(a, mu, sigma) Example: to find P(X < 4) enter: pnorm(4, 3, 0.75) Answer is P(X < 4) = 0.9088 To find P(X < a) with Excel, enter: =normdist(a,mu,sigma,true) ly Rom Stard Observe that P(X < a) + P(X > a) = 1, so P(X > a) = 1 P(X < a) Example: to find P(X > 4) = 1 P(X < 4) = 1 0.9088 = 0.0912 Example: to find P(X < 4) enter =normdist(1.5,3,0.75,true) To use your calculator, check the manual.
( How to find P(a < X < b) ( How to solve P(X < x) = p ly Rom Stard Observe that if a < b then P(X < a) + P(a < X < b) + P(X > b) = 1 so P(a < X < b) = 1 P(X < a) P(X > b) = 1 P(X < a) (1 P(X < b)) = P(X < b) P(X < a) P(2 < X < 4) = P(X < 4) P(X < 2) = 0.9088 0.0912 = 0.8176 ly Rom Stard If 0 < p < 1 we can find x such that P(X < x) = p With R, enter: qnorm(p,mu,sigma) For example, to find x such that P(X < x) = 0.9, enter: qnorm(0.9,3,0.75) Answer is x = 3.96 With Excel, enter: =norminv(p,mu,sigma) For example, to find x such that P(X < x) = 0.9, enter: =norminv(0.9,3,0.75) ( The stard normal variable Z ( Using table to find P(Z < a) ly Rom ly Rom Stard Statisticians always let Z st for the normal rom variable with mean µ = 0 stard deviation sigma = 1. Z is called the stard normal rom variable Stard To find P(Z < a) with the table, find a on the margin of the table read P from the inside Example: to find P(Z < 1.5) = P(X < 1.50) find 1.5 on th left margin 0 on the top. Read off 0.93319
( P(Z > a), etc. ( Solve P(Z < z) = p ly Rom Stard Just like before: P(Z > a) = 1 P(Z < a) P(a < Z < b) = P(Z < b) P(Z < a) ly Rom Stard If 0 < p < 1 we can find z such that P(Z < z) = p Find the number closest to p in the middle of the table. The marginal values give z. Example: solve P(Z < z) = 0.9 0.9 is between two numbers in table: 0.89973 0.90147 Number in table closest to 0.9 is 0.89973 Marginal value is z = 1.28 ( Find P(X < a) with tables ( Solve P(X < x) = p with tables ly Rom Stard Suppose X is a normal rom variable with mean µ = 3 stard deviation σ = 0.75. Find P(X < a) Let a = a µ σ. Then P(X < a) = P(Z < a ) a is called the z-score of a (with X in the backgroud) Evaluate P(X < 4) ( P(X < 4) = P Z < 4 3 ) = P(Z < 1.33) = 0.9082 0.75 Above the answer was P(X < 4) = 0.9088 Difference is 1.33 is not exactly 4 3 0.75 = 4 3 = 11 3 ly Rom Stard Suppose X is a normal rom variable with mean µ = 3 stard deviation σ = 0.75. If 0 < p < 1 solve P(X < x) = p First solve P(Z < z) = p Then let x = µ + σz Solve P(X < x) = 0.9. Above we saw athat P(Z < z) = 0.9 has solution z = 1.28 x = 3 + 0.75 1.28 = 3.96
( ly Rom Stard Suppose X is a normal rom variable modeling a characteristic of a population P(X < a) is the proportion of individuals in the population that have variable value less than a For example, if X represents heights, then P(X < 67) is the proportion of people that are less than 67" tall. We could also say P(X < a) is the probability that a romly selected individual has variable value less than a For example, if X represents heights, then P(X < 67) is the probability that a rom person is less than 67" tall. P(X > 67) is the probability that a rom person is more than 67" tall. P(67 < X < 68) is the probability that a rom person is between 67" 68" tall.