GM.Normal Probability Session 6&7

Continuous Probability Distribution - Normal Probability : 

Continuous Probability Distribution - Normal Probability Business Statistics - 2010 Dr. Gunjan Malhotra Assistant Professor Institute of Management Technology, Ghaziabad gmalhotra@imt.edu mailforgunjan@gmail.com 1 12/2/2011

Continuous Probability Distributions: 

Continuous Probability Distributions • A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. • Continuous probability distributions are described by: – The interval of possible values for the variable x. – The probability density f(x) associated with variable x. • The probability density is not easily interpreted, but the area under the probability density is a probability. 2 12/2/2011

Relative frequency on a histogram and probability density on a normal curve : 

Relative frequency on a histogram and probability density on a normal curve 3 12/2/2011

Normal Distribution/ Gaussian distribution: 

Normal Distribution/ Gaussian distribution ‘ Abrham De Moivre (1667-1754); Gauss (1809) & Laplace (1812) “Bell Shaped’ Symmetrical Mean, Median and Mode are Equal Location is determined by the mean, μ Spread is determined by the standard deviation, σ The random variable has an infinite theoretical range: +  to   Mean = Median = Mode X f(X) μ σ 4 12/2/2011

Using the standard normal probability table : 

Using the standard normal probability table z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359 .1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753 .2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 .3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517 .4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 .5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224 .6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549 .7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852 .8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133 .9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389 5 12/2/2011

Using the standard normal probability tables : 

Using the standard normal probability tables • What is the probability that a standard normal variable is between 0.60 and 1.34? • This area is given in the tables P(0.60 <  z  < 1.34) = 6 12/2/2011

Using the standard normal probability tables : 

Using the standard normal probability tables • What is the probability that a standard normal variable will be more than -1.88? • The trick is to notice that the normal distribution is symmetric. P(-1.88 < z)= 7 12/2/2011

Using the standard normal probability tables : 

Using the standard normal probability tables What is the probability that a standard normal variable is less than 0.88? • This area is given in the tables P(0< z<0.88)= 8 12/2/2011

Using the standard normal probability tables : 

Using the standard normal probability tables What is the probability that a standard normal variable will be between -1.88 and 0.88? • We break down the interval around zero. P(- 1.88 < z< 0.88 )=P(- 1.88 <z<0)+P(0<z< 0.88 )= 9 12/2/2011

Find the following probabilities and z-values : 

Find the following probabilities and z-values • P(0<z<2.45)= • P(z>2.45)= • P(z>-0.68)= • P(0.34<z<1.34)= 10 12/2/2011

Translation to the Standardized Normal Distribution: 

Translation to the Standardized Normal Distribution Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z) Translate from X to the standardized normal - the “Z” distribution: The Z distribution always has mean = 0 and standard deviation = 1 11 12/2/2011

Example: 

Example If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100. 12 12/2/2011

Comparing X and Z units: 

Comparing X and Z units Z 100 2.0 0 200 X ( μ = 100, σ = 50) ( μ = 0, σ = 1) 13 12/2/2011

PowerPoint Presentation: 

Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6) Z 0.12 0 X 8.6 8 μ = 8 σ = 10 μ = 0 σ = 1 Example- Normal Probabilities P(X < 8.6) P(Z < 0.12) 14 12/2/2011

Solution: Finding P(Z < 0.12): 

Z 0.12 Z .00 .01 0.0 .5000 .5040 .5080 .5398 .5438 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 Solution: Finding P(Z < 0.12 ) .5478 .02 0.1 . 5478 Standardized Normal Probability Table (Portion) 0.00 = P( Z < 0.12) P( X < 8.6) 15 12/2/2011

Finding the X value for a Known Probability: 

Steps to find the X value for a known probability: 1. Find the Z value for the known probability 2. Convert to X units using the formula: Finding the X value for a Known Probability 16 12/2/2011

Finding the X value for a Known Probability: 

Finding the X value for a Known Probability Example: Suppose X is normal with mean 8.0 and standard deviation 5.0. Now find the X value so that only 20% of all values are below this X X ? 8.0 0.2000 Z ? 0 17 12/2/2011

Find the Z value for 20% in the Lower Tail: 

Find the Z value for 20% in the Lower Tail 20% area in the lower tail is consistent with a Z value of -0.84 X ? 8.0 0.2000 Z -0.84 0 1. Find the Z value for the known probability 18 12/2/2011

Finding the X value: 

2. Convert to X units using the formula: Finding the X value So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 X 3.80 8.0 0.2000 Z -0.84 0 19 12/2/2011

Practice Questions: 

Practice Questions Lind – Pg – 233 – Q 10, 12 Pg 237 – Q 16 Pg 239 – Q 20, 22 Pg 241 – Q 24, 26 Pg 248 – Q 42, 46 Pg 250 – Q 58, 62 20 12/2/2011