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THE NORMAL DISTRIBUTION. NORMAL DISTRIBUTION Frequently called the “bell shaped curve” Important because –Many phenomena naturally have a “bell shaped”

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Presentation on theme: "THE NORMAL DISTRIBUTION. NORMAL DISTRIBUTION Frequently called the “bell shaped curve” Important because –Many phenomena naturally have a “bell shaped”"— Presentation transcript:

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 μ σ

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 “Standardizing” a Normal Random Variable X to Z 0 Z σ Z = 1 0 Z μ X σX = σσX = σσX = σσX = σ x0x0x0x0 Standardization preserves probabilities P(X>x 0 ) = P(Z>z) z

8 Representing X and Z on the Same Graph Both the x-scale and the corresponding z-scale can be represented on the same graph μ X σX = σσX = σσX = σσX = σ x0x0x0x0 0 Z z Example: Suppose μ = 10, σ =2. Illustrate P(X > 14). 10 σ =2 14 14 10 2 2 Thus P(X > 14) = P(Z>2)

9 Facts About the Normal 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 Example X is normally distributed with  = 244,  = 25 Find P(X < 200) For x = 200, z = (200-244)/25 = -1.76 X 244 σ = 25 200? 0 Z 0 Z -1.76

13 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 P (Z<-1.76)

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

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

18 Probability a Flight Takes Between 320 and 350 Minutes  = 20 320 X 350 0 Z 0 Z 1.50.9332.5000.9332 -.5000=.4332

19 Probability a Flight Takes Between 325 and 355 Minutes  = 20 320 X 355 325 0 Z 0 Z 1.75 0.25.9599.5987.9599 -.5987=.3612

20 Probability a Flight Takes Between 308 and 347 Minutes  = 20 320 X 308 347 0 Z 0 Z 1.35 -0.60.9115.2743.9115 -.2743=.6372

21 Probability a Flight Takes Between 275 and 285 Minutes  = 20 320 X 285 275 0 Z 0 Z -2.25 -1.75.0401.0122.0401 -.0122=.0279

22 Using Excel to Calculate Normal Probabilities Given values for μ and σ, cumulative probabilities P(X < x 0 ) are given by: Note that =NORMDIST(x 0,μ,σ,FALSE) returns the value of the density function at x 0, not a probability. If the value of z is given, then the cumulative probabilities P(Z<z) are given by: =NORMDIST(x 0, 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) =NORMSDIST(-1.75)-NORMSDIST(-2.25) =1-NORMSDIST(.75) =NORMSDIST(1.50)-NORMSDIST(0)=NORMSDIST(1.75)-NORMSDIST(.25) =NORMSDIST(1.35)-NORMSDIST(-.60) 0.027835

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

26 The Middle 75% of the Flight Times Lie Between What Two Values? Required to find x L and x U such that.7500 lies between x L and x U -- this means.1250 lies below x L and.1250 lies above x U (.8750 lies below x U )  = 20 320 X.7500 xLxLxLxL xUxUxUxU.1250 1 -.7500 =.2500 split between tails 0 Z 0 Z zLzLzLzL zUzUzUzU z L puts.1250 to leftz U puts.8750 to left -1.15 1.15 x L = 320+(-1.15)(20) 297 = 297 297 x U = 320+(1.15)(20) 343 = 343 343

27 Using Excel to Calculate x and z Values From Normal Probabilities Given values for μ and σ, the value of x 0 such that P(X < x 0 ) = 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(1-.1250,320,20) =NORMINV(.1250,320,20)=NORMINV(.1000,320,20) =NORMSINV(.1000)=NORMSINV(.1250) =NORMSINV(1-.1250)

29 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  = 20 μ X μ X 330.8000 0 Z Look up.8000 in the middle of the z-table.84 μ = 330 -.84(20) = 313.2 313.2

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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