7 THE NORMAL DISTRIBUTION STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 276 Learning Outcomes STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 277 At the end of this chapter you should be able to: 1. understand the properties of a normal random variable; 2. compute probabilities for a normal distribution using the standard normal distribution tables and also R; 3. determine the distribution of sums of normal random variables; 4. model and solve problems using normal random variables; 5. be able to model and solve problems using a combination of continuous and discrete random variables as appropriate. Contents STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 278 7.1 Introduction 7.2 Sums of Normal random variables 7.1 Introduction STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 279 A very important continuous distribution. Let the random variable X have a normal distribution with mean μ and variance σ 2. We write X ～ N(μ, σ2). Note that the variance is given in this description. Let Y = aX + b. Then E(Y) = aμ + b, Var(Y) = a2σ2, and Y also has a normal distribution, Y ～ N(aμ + b, a2σ2). Standardisation STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 280 In particular, if Z = X μ σ = 1 σ X μ σ ( a = 1 σ , b = μ σ ) then E(Z) = 0 and Var(Z) = 1, so Z ～ N(0, 1). We call Z the standard normal distribution. If Z ～ N(0, 1) and X = μ + σZ, then X ～ N(μ, σ2). Probability density function — effect of changing the mean STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 281 Below are sketched the pdf of the normal distribution for different values of the mean. Note the effect of changing the mean. 4 2 0 2 4 0 . 0 0 . 1 0 . 2 0 . 3 0 . 4 Changing the mean of a normal distribution x value D e n s i t y Means mu=1 mu= 1 mu=0 Probability density function — effect of changing the variance STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 282 Below are sketched the pdf of the normal distribution for different values of the variance. Note the effect of changing the variance. 6 4 2 0 2 4 6 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 Changing the standard deviations of a normal distribution x value D e n s i t y Standard Deviations sd=2 sd=0.5 sd=1 N(0, 1) distribution tables STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 283 Tables list cumulative probabilities for the standard normal distribution. Tables in exams will be the same as in the textbook, and also available online. Normal Distribution Problems STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 284 For X ～ N(μ, σ2), there are usually two types of problems. Given an interval for X, find the probability. Given a probability, find the value(s) of x (the inverse problem). The examples below illustrate the ideas. Example 7.1 STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 285 Let Z ～ N(0, 1). Determine the following. (i) P(0 < Z < 1.0) (ii) P(Z < 1.0) Example 7.1 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 286 (iii) P(Z > 1.0) (iv) P(Z < 2.51) Example 7.1 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 287 (iv) P(Z > 2.51) (vi) P(|Z| < 1.96) Example 7.1 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 288 (vii) P( 1.5 < Z < 2.7) (viii) The value of z such that P(Z < z) = 0.9505 Example 7.1 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 289 (ix) The value of z such that P(|Z| < z) = 0.95 Example 7.2 STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 290 Let X ～ N(5, 16). Determine the following. (i) P(X < 0) Example 7.2 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 291 (ii) P(X > 10) (iii) P( 5 ≤ X ≤ 7) Example 7.2 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 292 (iv) The value of c such that P(|X μ| < c) = 0.95 Example 7.3 STAT1520: Economic and Business Sta
tistics M Firth, R N Khan and B A Turlach – 293 Let X ～ N(μ, 0.25), and suppose P(X < 5.1) = 0.9772. Find the value of μ. Example 7.4 STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 294 A machine fills bottles of soft drink to a mean volume of 210 mL with a standard deviation of 10 mL. The label on the bottle specifies a volume of 200 mL. A bottle is under-filled if it contains less than the labelled volume. Assume that the volumes of the bottles are normally distributed. (a) What percentage of bottles are under-filled Example 7.4 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 295 (b) In order to reduce the percentage of under-filled bottles to 1% the company decides to adjust the standard deviation of the volumes filled by the machine. What should the standard deviation be reduced to 7.2 SUM OF INDEPENDENT NORMAL RANDOM VARIABLES STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 296 Let X ～ N(μX, σ 2 X ), Y ～ N(μY, σ 2 Y ), with X and Y independent. Then put W = X ± Y . Then W ～ N(μW, σ 2 W ), where μW = μX ± μY and σ2W = Var(X ± Y) = Var(X) + Var(Y) = σ2X + σ 2 Y. This result can be extended to a sum of several normal random variables. Result STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 297 Let X1, X2, · · · , Xn be independent normal random variables with mean E(Xi) = μi and Var(Xi) = σ 2 i , i = 1, 2, · · · , n. Put Y = n∑ i=1 Xi. Then Y ～ N(μY , σ 2 Y ), where μY = E n∑ i=1 Xi = n∑ i=1 E (Xi) = n∑ i=1 μi and σ 2 Y = Var n∑ i=1 Xi = n∑ i=1 Var (Xi) = n∑ i=1 σ 2 i . 7.2b SUM OF NON INDEPENDENT NORMAL RANDOM VARIABLES STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 298 Let X ～ N(μX, σ 2 X ), Y ～ N(μY, σ 2 Y ), and put W = X ± Y . Then μW = μX ± μY and σ2W = Var(X ± Y) = Var(X) + Var(Y) ± 2Cov(X,Y). Furthermore W ～ N(μW, σ 2 W ) if X and Y are what we call bivariate normal. While not every pair of normal random variables are bivariate normal, we will assume that to be the case in STAT1520 without needing to mention it. So if you are working on any problem with the sum of normal random variables, you may assume that the sum is normally distributed. Example 7.5 STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 299 Suppose the income of executives is normally distributed with the incomes of men and women independent of each other. The means and standard deviations (in $10,000) respectively are 20, 2.5 for men and 15, 2.0 for women. (a) What is the probability that a randomly chosen female executive earns more than a randomly chosen male executive Example 7.5 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 300 Example 7.5 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 301 (b) Find the probability that a randomly chosen female executive and a randomly chosen male executive have a combined income of more than $400,000. Example 7.6 STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 302 A machine makes washers with hole diameters that are normally distributed, with mean 15.2 mm and variance 0.03 mm2. Another machine makes bolts with diameters that are normally distributed, with mean 15.0 mm and variance 0.01 mm2. (a) What is the probability that a randomly selected bolt with fit through a randomly selected washer Example 7.6 (ctd) STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 303 (b) What should be the mean diameter of the washer holes if 99% of the bolts are to fit the washers Critical values of the standard normal distribution STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 304 We often need the value of the Z-distribution corresponding to given tail probabilities. For convenience the commonly used values are listed below the standard normal table. They list the Z-values corresponding to either one-tail to two-tail probabilities. Example Find the value of z such that 1. P(Z > z) = 0.01 2. P(Z < z = 0.05) 3. P(|Z| > z = 0.05) 4. P(|Z| > z = 0.01)