LESSON 10: NORMAL DISTRIBUTION

Transcription

1 LESSON 10: NORMAL DISTRIBUTION Outline Normal distribution Area under the curve, probability, percentile value Given z find area Given percentile value find z Given x find area Given percentile value find x 1 NORMAL DISTRIBUTION THE PROBABILITY DENSITY FUNCTION If a random variable X with mean µ and standard deviation σ is normally distributed, then its probability density function is given by f 1 σ ( 1/2 ) ( x µ ) ( x) = e σ 2π [ / ] 2 2 1

2 NORMAL DISTRIBUTION THE PROBABILITY DENSITY FUNCTION f (x ) Area under the curve = 1.00 f (x ) Mean, m=50 SD, s=10 Area between the vertical lines = P (40 X 60) x - VALUES 3 NORMAL DISTRIBUTION EFFECT OF CHANGING STANDARD DEVIATION SD,s=10 SD,s=15 SD,s=20 Mean,m= x-values 4 2

3 NORMAL DISTRIBUTION EFFECT OF CHANGING MEAN SD,s= x -values 5 f (x) Area under the curve = z -VALUES Mean=0 SD=1 6 3

4 Area = 1.00 Area = 0.50 Area = RELATIONSHIP BETWEEN x AND z If a random variable X is normally distributed with mean µ and standard deviation σ, then µ=50 σ=10 x µ z = σ or, x = µ + zσ z= x=

5 TABLE, z-values, AREA AND PROBABILITY Example 1.1: Table D, Appendix A, pp shows the area under the curve from Z=-8 to some z value. For example, the area from Z=-8 to Z= =1.34 is So, Φ 1.34 = , P Z 1.34 = P Z 1.34 = 0. ( ) ( ) ( ) TABLE, z-values, AREA AND PROBABILITY Example 1.2: The area shown on the table can be used to get many other areas. For example, using the fact that the area under the curve is 1.0, the area from Z=1.34 to Z= is = So, P( 1.34 Z ) = P( Z 1.34) =

6 TABLE, z-values, AREA AND PROBABILITY Example 1.3: The area shown on the table can be used to get area between any two z-values. For example, the area from z 1 =-1.25 to z 2 =1.34 is =0.8043, where is the area obtained from Table D for z 1 = So, P 1.25 Z 1.34 = 0. ( ) GIVEN z, FIND PROBABILITY Example 2: Find the following: 1. P 2. P 3. P 4. P ( Z 1.63) ( Z 1.63) ( 1.05 Z 1.63) ( 1.05 Z 1.63) 5. P 6. P ( Z < 1.63) ( Z = 1.63) 12 6

7 GIVEN z, FIND PROBABILITY : EXCEL Excel function NORMSDIST(z) provides the area under the standard normal distribution curve on the left side of z. Example: NORMSDIST(1.34) = = P Z 1.34 = Φ 1.34 ( ) ( ) P ( Z 1.34) =? Area =? To get the area on the left of Z = 1.34, F(1.34), use Excel function NORMSDIST(1.34). z= AREA AND PERCENTILE A percentile is the value at or below which the stated percentage of units lie. Therefore, percentile corresponds to an area under the curve. For example, if GMAT scores are normally distributed with the 78 th percentile 600, then 78% scores are less than 600 and P( X 600 ) = Then, the area on the left of X = 600 is If the 78 th percentile is 600, then the area on the left of X=600 is Area = 0.78 X=

8 GIVEN AREA OR PERCENTILE, FIND z Example 3: If return on investment of a fund has a mean 0 and standard deviation 1, find the returns that corresponds to following percentiles: th th 15 GIVEN AREA OR PERCENTILE, FIND z : EXCEL Excel function NORMSINV(p) provides the value of z corresponding to the 100p th percentile. For example, NORMSINV(0.33)= So, for the standard normal distribution, the 33 rd percentile is Area =0.33 z=? To get the z value for which area on the left, F(z) = 0.33, use Excel function NORMSINV(0.33). 16 8

9 NORMAL DISTRIBUTION GIVEN x, FIND PROBABILITY Example 4.1: A retailer has observed that the monthly demand of an item is normally distributed with a mean of 650 and standard deviation of 50 units. What is the probability that the demand of the item in the next month will not exceed 700 units? 1. Compute z P( X 700 ) =? 2. Find area from the Table 3. Find probability 17 Example 4.2: A retailer has observed that the monthly demand of an item is normally distributed with a mean of 650 and standard deviation of 50 units. What is the probability that the demand of the item in the next month will exceed 600 units? 1. Compute z NORMAL DISTRIBUTION GIVEN x, FIND PROBABILITY P( X 600 ) =? 2. Find area from the Table 3. Find probability 18 9

10 NORMAL DISTRIBUTION GIVEN x, FIND PROBABILITY Example 4.3: A retailer has observed that the monthly demand of an item is normally distributed with a mean of 650 and standard deviation of 50 units. What is the probability that the demand of the item in the next month will be between 600 and 700 units? 1. Compute z 1 and z P 2 ( 600 X 700) =? 2. Find areas from the Table 3. Find probability 19 NORMAL DISTRIBUTION GIVEN x, FIND PROBABILITY : EXCEL Excel function NORMDIST(x,µ,σ,TRUE) provides the area under the standard normal distribution curve on the left side of x. For example, NORMDIST(700,650,50,TRUE) = P( X 700 ) =? To get the area on the left of X = 700, when µ=600, s=50, use Excel function NORMSDIST(700,650,50,TRUE). Area=? µ=600 s=50 X =

11 NORMAL DISTRIBUTION GIVEN AREA OR PERCENTILE, FIND x Example 5: A retailer has observed that the monthly demand of an item is normally distributed with a mean of 650 and standard deviation of 50 units. If the retailer wants to meet demand with probability 0.90, how many units should be ordered for the next month? Assume that there is no units in the inventory. 1. Find z from the table 2. Find x 21 GIVEN AREA OR PERCENTILE, FIND x : EXCEL Excel function NORMINV(p,µ,σ) provides 100p th percentile when mean is µ and standard deviation σ. So, the function also gives that value of X for which the area on the left side of X is p. For example, NORMINV(0.90,650,50) = 714. So, the 90 th percentile is 714. To get the x value for which area on the left = 0.90, use Excel function NORMSINV(0.90,650,50) if mean is 650 and standard deviation 50. Area = 0.90 X =? µ=600 s=

12 READING AND EXERCISES Lesson 10 Reading: Section 7-4, pp Exercises: 7-35, 7-36,