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Chapter 6: Normal Distributions 6.1 Normal Distribution Learning Activity 6.1-1 Plotting a normal curve Open z-plot.xls Click on the chart to see the source data Click in one of the cells in column C. See Excel formula. f ( y ) 1 exp( 1 ( y )2 ) Z (Y ) / f ( z) 1 exp( z 2 ) 2 2 2 Learning Activity 6.1-2 Practice with z-values Open z-demo.xls!start. Calculate the mean and s.d of the X values In column C calculate the z-value for each of the X values. Calculate the mean and s.d. of the z-values. Click on the z-demo worksheet tab. This worksheet generates random numbers for the X values. z-values: z= (X – mean)/s.d. 2 Example 1 IQ scores are close to normally distributed and standardized to have mean = 100 and a standard deviation of = 16. If you take an IQ test and get a score of 125, what is your percentile, or what proportion of the population would be below you? Calculate z = (125 - 100)/16 = 1.56 Look up Tables.xls!Normal +z, and you find 0.94. That is you are at the 94th percentile, or 94% of the population would be at or below your IQ score of 125. The normal probabilities are valid only to which population is normal. Do a frequency distribution to see if it looks normal. If a distribution has a peak in the middle and sort of tapers down at both ends, it probably will be called approximately normal. Benchmark z-values The z-values corresponding to 1%, 5%, and 10%. Probability one-tail two-tail 0.1/.9 + or -1.28 + and -1.645 0.05/.95 + or -1.645 + and -1.96 0.01/.99 + or -2.33 + and -2.58 (See Tables.xls!Benchmark_z_values.) Learning Activity 6.1-5 Important z-values Verify these values by (1) Normal curve table (2) Excel function NORMSINV(p) to find the z value corresponding to P(Z < z) = p. (3) MegaStat | Probability | Normal Distribution Empirical Rule What percent of a normal distribution falls within 1 s.d. of the mean? That is the probability between z = -1 and z = +1. From Tables.xls, P(Z < 1) – P(Z < -1) = 0.8413-0.1587 = 0.6827 (68%) Similarly, the probability within 2 s.d. of the mean is P(Z < 2) – P(Z < -2) = 0.9772-0.0228 = 0.9544 (95%) The probability with 3 s.d. of the mean is 0.9973. Use MegaStat|Probability|Normal Distribution to show the following figues. (Mark Calcculate z given P, shading: upper/lower, Color: Transparent, and Overlay) Approximately 68% of the area is within +/-1 Approximately 95% of the area is within +/-2 Almost all of the area is within +/-3 Calculate the values below using the three methods: Normal curve table Excel functions: NORMSDIST(z), or NORMDIST(X,mean,stdev,1) MegaStat | Probability | Normal Distribution Learning Activity 6.A-1 Practice with the normal distribution Assume an examination has a mean of 84.2 and a standard deviation of 10.9. What is the probability of being above 100? What is the probability of being between 77 and 97? If an examination score has to be at the 94th percentile to be an A, what score would that be? See Normal_practice.xls Relationship between the normal and binomial distributions As n approaches infinity, the binomial distributions are mathematically equivalent. Plot (1) the binomial probabilities with n = 1000 and p = 0.2 (2) the binomial probabilities with n = 1000 and p = 0.5 Binomial distribution (n = 1000, p = 0.2) Binomial distribution (n = 1000, p = 0.5) 0.04 0.03 0.03 0.03 0.02 P(X) 0.02 0.02 0.02 X X 571 562 553 544 535 526 517 508 499 490 481 472 463 454 445 436 262 255 248 241 234 227 220 213 206 199 192 185 178 171 0.00 164 0.00 157 0.01 150 0.01 427 0.01 0.01 143 P(X) 0.03 Example 2 If a university wanted to accept students at the 86th percentile on an entrance examination that had a mean of 400 and a standard deviation of 60, to what score would that correspond? From tables.xls!Normal +z, we see that P(Z < 1.08) = 0.8599 Hence, a z-value of 1.08 corresponds to 1.08*60 + 400 = 464.8 Using Excel Functions To calculate normal probability NORMSDIST(z) NORMDIST(Y, mean, stdev, true/false) To calculate z or Y given a probability NORMSINV(p) NORMINV(p, mean, stdev) NORMINV(0.86, 400,60) = 464.8