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Mean,Mean. A FormulaA Formula: page 6 The Sample Sample If X is any random variable, then, as n increases without bound, x X the distribution of its standardized sample mean, , approaches the X n distribution of the standard normal random variable, Z, whose p.d.f. is 1 0.5z 2 f Z ( z) e . 2 This is called the Central Limit Theorem. (material continues) T C I Normal Distributions. Standard Normal Normal Distributions, Standard Normal 1. STANDARD NORMAL(Mean 0 & Standard deviation 1) In The Sample Mean we derived the probability density function 1 0.5z 2 f Z ( z) e 2 for the standard normal random variable Z. We can use integration and fZ to compute probabilities for Z. Example 1. Compute P(0.74 Z 1.29). As we saw in Integration, 1.29 P( 0.74 Z 1.29) 0.74 1 0.5z 2 e dz 2 To evaluate the integral open Integrating.xls and enter the function as =(1/SQRT(2*PI()))*EXP(-0.5*x^2). Recall that x is the only variable that can be used in Integrating.xls. Integrating.xls (material continues) T C I Normal Distributions. General Normal Normal, General Normal 2. GENERAL NORMAL(Any Mean & Any Standard deviation/But its standardization is a standard normal distribution) The adjective “standard”, used in standard normal distributions, implies that there are “non-standard” normal distributions. This is indeed the case. A random variable, X, is called normal if its standardization, X X S , X has a standard normal distribution. It can be shown that the probability density function for a normal random variable, X, with mean X and standard deviation X has the following form. 2 fX ( x) 1 X 2 (material continues) x X 0.5 X e T C I Normal Distributions• Standard Normal Random Variable (Z) f z e Z • p.d.f. 2 1 0.5 z 2 • Can use Integrating.xls to find probabilities Normal Distributions• Ex1. Find P 1.34 Z 0.68 • Soln: show ex1 excel file P 1.34 Z 0.68 0.68 1 1.34 2 0.6616 e 0.5 z 2 dz Normal Distributions • Ex. Find a number z 0 so that PZ z 0 0.9750. • Soln:show ex2 excel file z0 1 0.5 z 2 PZ z 0 e dz 2 z 0 1.9600 Normal Distributions • The previous example tells us that 97.5% of all data for a standard normal random variable lies in the interval . , 1.96 • This means that 2.5% of the data lies above z = 1.96 • Graphically, we have the following: Normal Distributions • The shaded region corresponds to 97.5% of all possible area (note 2.5% is not shaded) -6 -5 -4 -3 -2 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -1 0 1 2 3 4 5 6 1.96 Normal Distributions • Due to symmetry, we get 95% of the area shaded with 5% not shaded (2.5% on each side) -6 -1.96 -5 -4 -3 -2 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -1 0 1 2 3 4 5 6 1.96 Normal Distributions • This means that a 95% confidence interval for the standard normal random variable Z is (-1.96, 1.96) -1.96 -6 -5 -4 -3 -2 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -1 0 1 2 3 4 5 6 1.96 Normal Distributions • A 95% confidence interval tells you how well a particular value compares to known data or sample data • The interval that is constructed tells you that there is a 95% probability that the interval will contain the mean of X. • Another interpretation is that 95% of all values found in a sample should lie within this 95% confidence interval. Normal Distributions • Possible formulas: Z Standard Normal random variable P 1.96 Z 1.96 0.95 Normal Distributions • Possible formulas: x x P 1.96 1.96 0.95 x x x P 1.96 s 1.96 0.95 n Important • Possible formulas: S If is X X X X x x x x x X n unknown The sample standard deviation, s ( sm all ) , will be a very good approximation for X Normal Distributions • Remember, that -1.96 and 1.96 were special values that apply to a 95% confidence interval • You need to find different values for other types of confidence intervals. • Ex. Find a 99% confidence interval for Z. Find a 90% confidence interval for Z. Normal Distributions • Soln:PZ z1 0.9950 PZ z 2 0.9500 z1 2.5759 z 2 1.6449 • Ex. What is the confidence interval for Z with 1 standard deviation? 2 standard deviations? 3 standard deviations? Normal Distributions • Soln: P 1 Z 1 0.6827 -6 -5 -4 -3 -2 P 2 Z 2 0.9545 -6 -5 -4 -3 -2 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -1 0 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -1 0 1 1 2 2 3 3 4 4 5 6 5 6 Normal Distributions • Soln: P 3 Z 3 0.9973 -6 -5 -4 -3 -2 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -1 0 1 2 3 4 5 6 Normal Distributions • Since Z is a standard normal random variable, Z would have standardized some variable X. • So, Z X X X Normal Distributions • Ex. Suppose X is a normal random variable with and X 15. Find a 95% X 120 confidence interval for X if the 95% confidence interval for Z is (-1.96, 1.96). Normal Distributions • Soln: P 1.96 Z 1.96 0.95 X X P 1.96 1.96 0.95 X X 120 P 1.96 1.96 0.95 15 P 1.96 15 X 120 1.96 15 0.95 P120 1.96 15 X 120 1.96 15 0.95 P90.6 X 149.4 0.95 So, the 95% confidence interval for X is (90.6, 149.4). We are 95% confident that this interval contains the true mean Normal Distributions • Ex. Suppose X is a normal random variable. If a sample of size 34 was taken with x 120 and , find a 95% confidence interval for s 15 the sample mean-remember this is x if the 95% confidence interval for Z is (-1.96, 1.96). Normal Distributions P 1.96 Z 1.96 0.95 • Soln: x x P 1.96 1.96 0.95 x x x P 1.96 1.96 0.95 X n x x P 1.96 1.96 0.95 s n x 120 P 1.96 1.96 0.95 15 34 Normal Distributions • Soln: 15 15 P 1.96 x 120 1.96 0.95 34 34 15 15 P120 1.96 x 120 1.96 0.95 34 34 P114.9579 x 125.0421 0.95 So, the 95% confidence interval for x is (114.9579, 125.0421). We are 95% confident that this interval contains the true mean Normal Distributions • General Normal Random Variable • p.d.f. f X x 1 X 2 e x X 0.5 X 2 • Probabilities are done similarly to Standard NRV Normal Distributions • Ex. If X is a normal random variable representing exam 1 scores with mean 75 and standard deviation 10, find P70 X 100 . • Soln: P70 X 100 100 1 10 2 0.6853 70 e x 75 0.5 10 2 dx Normal Distributions • NORMDIST function in Excel • Can calculate p.d.f. and c.d.f. values for a normal random variable • Ex. If X is a normal random variable representing exam 1 scores with mean 75 and standard deviation 10, find P70 X 100. Normal Distributions(show excel) • Soln: P70 X 100 FX 100 FX 70 NORMDIST 100, 75, 10, TRUE NORMDIST 70, 75, 10, TRUE 0.6853 100 75 10 TRUE 70 75 10 TRUE Normal Distributions • Specific values for the p.d.f. can also be calculated using NORMDIST • Ex. Find height of p.d.f. of a normal random variable X at X = 90 that has a normal distribution with 71 and X 12 . X Normal Distributions • Soln: Two ways to solve=0.009 1 (1) Evaluate 12 2 e 90 71 0.5 12 2 (2) Evaluate =NORMDIST(90, 71, 12, FALSE) using Excel Normal Distributions • How does the mean and standard deviation affect the shape of the Normal Random Variable graph? • Ex. Graph the p.d.f. of a normal random variable with the following characteristics: (1) X 0 and X 1 X 5 0 X (2) and 1 4 X X (3) and X 4 5 X (4) and Why? Normal Distributions Recall-General Normal random variable p.d.f • Soln: (1) X 0 Max Height X 1 and 0.40 The y-values of the graph around x = -3 and x = 3 are very small -6 -5 -4 -3 -2 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -1 0 1 2 3 4 5 6 Normal Distributions • General Normal Random Variable e^(0)=1 • p.d.f. f X x 1 X 2 x X 0.5 e X y int ercept f X 0 1 1 2 2 00 0.5 1 e 2 ~ 1 ~ 0.4 2 Normal Distributions-sec1-4/13 Why? • Soln: (2) X 0 and Max Height 0.08 (This is 0.40 std. dev.) Y values very small Around x = -15 and x = 15 -25 (This is 3 standard deviations from the mean- 3 * -20 -15 X) -10 X 5 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 -5 0 5 10 15 20 25 Normal Distributions • General Normal Random Variable e^(0)=1 • p.d.f. x X 0.5 X 00 0.5 5 2 1 f X x e X 2 f X 0 1 e 5 2 2 ~ 1 ~ 0.4 / 5 ~ 0.08 5 2 Normal Distributions • Soln: (3) X 4 Max Height 0.40 (At x = 4) Y values very small around x = 1 and x = 7 (This is 3 standard deviations from the mean) X 1 and -2 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -1 0 1 2 3 4 5 6 7 8 9 10 Normal Distributions • Soln: (4) X 4 Max Height 0.08 (This is 0.40 std. dev.) Y values very small around x = -11 and x = 19 (This is 3 standard deviations from the mean) and X 5 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 -26 -21 -16 -11 -6 -1 4 9 14 19 24 29 34 Normal Distributions • Ex. Find the mean and standard deviation for the following normal random variables graphed. (A) 0.14 0.12 0.10 0.08 0.06 0.04 0.02 -9 -6 0.00 -3 0 3 6 9 12 15 18 21 Normal Distributions 0.14 0.12 0.10 0.08 0.06 0.04 0.02 -9 -6 0.00 -3 0 3 6 9 12 15 18 21 • Mean is 6 and standard deviation is 3 Normal Distributions • (B) 0.25 0.20 0.15 0.10 0.05 -17 -15 -13 -11 -9 -7 -5 0.00 -1 -3 1 3 Normal Distributions 0.25 0.20 0.15 0.10 0.05 -17 -15 -13 -11 -9 -7 -5 0.00 -3 -1 1 3 • Mean is -7 and standard deviation is 2 Normal Distributions • (C) 0.009 0.008 0.007 0.006 0.005 0.004 0.003 0.002 0.001 0 50 100 150 200 250 300 350 400 450 500 550 Normal Distributions 0.009 0.008 0.007 0.006 0.005 0.004 0.003 0.002 0.001 0 50 100 150 200 250 300 350 400 450 500 550 • Mean is 300 and standard deviation is 50 Normal Distributions • Relationship between p.d.f. and c.d.f. • Pa X b FX b FX a • Pa X b a f X x dx b • So, FX b FX a b a f X x dx