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THE NORMAL DISTRIBUTION

THE NORMAL DISTRIBUTION. NORMAL DISTRIBUTION. Frequently called the “bell shaped curve” Important because Many phenomena naturally have a “bell shaped” distribution Normal distribution is the “limiting” distribution for many statistical tests. Characterizing a Normal Distribution.

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THE NORMAL DISTRIBUTION

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  1. THE NORMAL DISTRIBUTION

  2. NORMAL DISTRIBUTION • Frequently called the “bell shaped curve” • Important because • Many phenomena naturally have a “bell shaped” distribution • Normal distribution is the “limiting” distribution for many statistical tests

  3. Characterizing a Normal Distribution • To completely characterize a normal distribution, we need to know only 2 parameters: • It’s mean () • It’s standard deviation () • A normal distribution curve can be now drawn as follows:

  4. NORMAL PROBABILITY DENSITY CURVE σ μ

  5. NORMAL PROBABILITY DENSITY FUNCTION • The density function for a normally distributed random variable with mean μ and standard deviation σ is:

  6. CALCULATING NORMAL PROBABILITIES • There is no easy formula for integrating f(x) • However, tables have been created for a “standard” normal random variable, Z, which has  = 0 and σ =1 • Probabilities for any normal random variable, X, can then be found by converting the value of x to z where z = the number of standard deviations x is from its mean, 

  7. σX =σ μ X x0 P(X>x0) = P(Z>z) σZ = 1 Z Z 0 0 z “Standardizing” a Normal Random Variable X to Z Standardization preserves probabilities

  8. σX =σ μ X x0 14 z Representing X and Z on the Same Graph Both the x-scale and the corresponding z-scale can be represented on the same graph 10 14 2 σ =2 10 0 Z 2 Example: Suppose μ = 10, σ =2. Illustrate P(X > 14). Thus P(X > 14) = P(Z>2)

  9. Facts About theNormal Distribution • mean = median = mode • Distribution is symmetric • 50% of the probability is on each side of the mean • Almost all of the probability lies within 3 standard deviations from the mean • On the z-scale this means that almost all the probability lies in the interval from z = -3 to z = +3

  10. Using the Cumulative Normal (Cumulative z) Table • The cumulative z-table • Gives the probability of getting a value of z or less • P(Z < z) • Left-tail probabilities • Excel gives left-tail probabilities • To find any probability from a z-table: • Convert the problem into one involving only left-tail probabilities P(Z < a) = z-table value for a P(Z > a) = 1-(z-table value for a) P(a<Z<b) = (z-table value for b) – (z-table value for a)

  11. Using the Normal Table • Look up the z value to the first decimal place down the first column • Look up the second decimal place of the z-value in the first row • The number in the table gives the probability P(Z<z)

  12. σ = 25 ? X 244 Example • X is normally distributed with  = 244,  = 25 • Find P(X < 200) • For x = 200, z = (200-244)/25 = -1.76 200 -1.76 0 Z

  13. P (Z<-1.76) Using Cumulative Normal Tables z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 -3.0 .0013 .0013 .0013 .0012 .0160 .0012 .0011 .0011 .0010 .0010 -2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036 -2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048 -2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064 -2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 -2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 -2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 -2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 -1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 -1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 -1.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367 -1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455 -1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559

  14. EXAMPLE • Flight times from LAX to New York: • Are distributed normal • The average flight time is 320 minutes • The standard deviation is 20 minutes

  15. Probability a flight takes exactly 315 minutes • P(X = 315 ) = 0 • Since X is a continuous random variable

  16. From Table .1587  = 20 320 X Probability a Flight Takes Less Than 300 Minutes 300 -1.00 0 Z

  17. From Table .7734  = 20 1-.7734= .2266 320 X 335 Probability a Flight Takes Longer Than 335 Minutes .75 0 Z

  18. .9332 - .5000= .4332 .9332  = 20 .5000 320 X 350 Probability a Flight Takes Between 320 and 350 Minutes 1.50 0 Z

  19. .9599 - .5987= .3612 .9599  = 20 .5987 325 320 X 355 Probability a Flight Takes Between 325 and 355 Minutes 0.25 0 Z 1.75

  20. .9115 - .2743= .6372 .9115  = 20 .2743 320 X 308 347 Probability a Flight Takes Between 308 and 347 Minutes 1.35 -0.60 0 Z

  21. .0401 - .0122= .0279 .0401  = 20 .0122 320 X 275 285 Probability a Flight Takes Between 275 and 285 Minutes -2.25 -1.75 0 Z

  22. Using Excel to Calculate Normal Probabilities • Given values for μ and σ, cumulative probabilities P(X < x0) are given by: • Note that =NORMDIST(x0,μ,σ,FALSE) returns the value of the density function at x0, not a probability. • If the value of z is given, then the cumulative probabilities P(Z<z) are given by: =NORMDIST(x0, μ, σ, TRUE) =NORMSDIST(z)

  23. =NORMDIST(300,320,20,TRUE) =1-NORMDIST(335,320,20,TRUE) =NORMDIST(350,320,20,TRUE)-NORMDIST(320,320,20,TRUE) =NORMDIST(355,320,20,TRUE)-NORMDIST(325,320,20,TRUE) =NORMDIST(347,320,20,TRUE)-NORMDIST(308,320,20,TRUE) =NORMDIST(285,320,20,TRUE)-NORMDIST(275,320,20,TRUE) =NORMSDIST(-1.00) =1-NORMSDIST(.75) =NORMSDIST(1.50)-NORMSDIST(0) =NORMSDIST(1.75)-NORMSDIST(.25) =NORMSDIST(1.35)-NORMSDIST(-.60) =NORMSDIST(-1.75)-NORMSDIST(-2.25)

  24. Calculating x and z Values From Normal Probabilities • Basic Approach • Convert to a cumulative probabiltity • Locate that probability (or the closest to it) in the Cumulative Standard Normal Probability table • This gives the z value • This is the number of standard deviations x is from the mean x = μ + zσ Note: z can be a negative value

  25.  = 20 X = 320+(-1.28)(20) = 294.4 320 X .9000 .1000 294.4 ? -1.28 (approx.) 90% of the Flights Take At Least How Long? • .9000 of the probability lies above the x value • .1000 lies below the x value Look up .1000 in middle of z table 0 Z

  26.  = 20 1 - .7500 =.2500 split between tails xL= 320+(-1.15)(20) = 297 xU= 320+(1.15)(20) = 343 .7500 320 X .1250 .1250 xL xU 297 343 zU zL zLputs .1250 to left zU puts .8750 to left The Middle 75% of the Flight Times Lie Between What Two Values? • Required to find xL and xU such that .7500 lies between xL and xU -- this means .1250 lies below xL and .1250 lies above xU (.8750 lies below xU) 1.15 0 Z -1.15

  27. Using Excel to Calculate x and z Values From Normal Probabilities • Given values for μ and σ, the value of x0 such that P(X < x0) = p is given by: =NORMINV(p, μ, σ) • The value of z such that P(Z<z) = p is given by: =NORMSINV(p)

  28. =NORMINV(.1000,320,20) =NORMINV(.1250,320,20) =NORMINV(1-.1250,320,20) =NORMSINV(.1000) =NORMSINV(.1250) =NORMSINV(1-.1250)

  29. μ = 330 - .84(20) = 313.2  = 20 .8000 μ X 330 313.2 .84 What Would the Mean Have to Be So That 80% of the Flights Take Less Than 330 Minutes? • Since x = μ + zσ, then μ = x - z σ Look up .8000 in the middle of the z-table 0 Z

  30. REVIEW • Normal Distribution • Importance and Properties • Converting X to Z • Use of Tables to Calculate Probabilities • Use of Excel to Calculate Probabilities

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