Presentation on theme: "The Sample Mean. A Formula: page 6"— Presentation transcript:
1The Sample Mean. A Formula: page 6 If X is any random variable, then, as n increases without bound,the distribution of its standardized sample mean, approaches thedistribution of the standard normal random variable, Z, whose p.d.f. isThis is called the Central Limit Theorem.TCI(material continues)
2Normal Distributions, Standard Normal 1. STANDARD NORMAL(Mean 0 & Standard deviation 1)In The Sample Mean we derived the probability density functionfor 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,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.xlsTCI(material continues)
3Normal Distributions. General Normal Normal, General Normal2. 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,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.(material continues)TCI
4Normal Distributions- Standard Normal Random Variable (Z)p.d.f.Can use Integrating.xls to find probabilities
5Normal Distributions- Ex1. FindSoln: show ex1 excel file
6Normal Distributions Ex. Find a number so that . Soln:show ex2 excel file
7Normal DistributionsThe previous example tells us that 97.5% of all data for a standard normal random variable lies in the intervalThis means that 2.5% of the data lies above z = 1.96Graphically, we have the following:
8Normal DistributionsThe shaded region corresponds to 97.5% of all possible area (note 2.5% is not shaded)1.96
9Normal DistributionsDue to symmetry, we get 95% of the area shaded with 5% not shaded (2.5% on each side)
10Normal DistributionsThis means that a 95% confidence interval for the standard normal random variable Z is (-1.96, 1.96)
11Normal DistributionsA 95% confidence interval tells you how well a particular value compares to known data or sample dataThe 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.
12Normal Distributions Possible formulas: Z Standard Normal random variable
14Important Possible formulas: If is unknown The sample standard deviation,, will be a very good approximation for
15Normal DistributionsRemember, that and 1.96 were special values that apply to a 95% confidence intervalYou 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.
16Normal Distributions Soln: Ex. What is the confidence interval for Z with 1 standard deviation? 2 standard deviations? 3 standard deviations?
19Normal DistributionsSince Z is a standard normal random variable, Z would have standardized some variable X.So,
20Normal DistributionsEx. Suppose X is a normal random variable with and Find a 95% confidence interval for X if the 95% confidence interval for Z is (-1.96, 1.96).
21Normal Distributions Soln: So, the 95% confidence interval for X is (90.6, 149.4). We are 95% confident that this interval contains the true mean
22Normal DistributionsEx. Suppose X is a normal random variable. If a sample of size 34 was taken with and , find a 95% confidence interval for the sample mean-remember this isif the 95% confidence interval for Z is (-1.96, 1.96).
24Normal Distributions Soln: So, the 95% confidence interval for is ( , ).We are 95% confident that this interval contains the true mean
25Normal Distributions General Normal Random Variable p.d.f. Probabilities are done similarly to Standard NRV
26Normal DistributionsEx. If X is a normal random variable representing exam 1 scores with mean 75 and standard deviation 10, findSoln:
27Normal Distributions NORMDIST function in Excel Can calculate p.d.f. and c.d.f. values for a normal random variableEx. If X is a normal random variable representing exam 1 scores with mean 75 and standard deviation 10, find
29Normal DistributionsSpecific values for the p.d.f. can also be calculated using NORMDISTEx. Find height of p.d.f. of a normal random variable X at X = 90 that has a normal distribution with and
30Normal Distributions Soln: Two ways to solve=0.009 (1) Evaluate (2) Evaluate =NORMDIST(90, 71, 12, FALSE) using Excel
31Normal DistributionsHow 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) and(2) and(3) and(4) and
32Normal Distributions Soln: (1) and Max Height 0.40 Why?Recall-General Normal random variable p.d.fSoln: (1) andMax HeightThe y-values of the grapharound x = -3 and x = 3are very small
33Normal DistributionsGeneral Normal Random Variablep.d.f.e^(0)=1
34Normal Distributions-sec1-4/13 Why?Soln: (2) andMax Height(This is std. dev.)Y values very smallAroundx = -15 andx = 15(This is 3 standarddeviations from the mean )
35Normal DistributionsGeneral Normal Random Variablep.d.f.e^(0)=1
36Normal Distributions Soln: (3) and Max Height 0.40 (At x = 4) Y values very smallaround x = 1 and x = 7(This is 3 standarddeviations fromthe mean)
37Normal Distributions Soln: (4) and Max Height 0.08 (This is std. dev.)Y values very smallaround x = -11 and x = 19(This is 3 standarddeviations from themean)
38Normal DistributionsEx. Find the mean and standard deviation for the following normal random variables graphed.(A)
39Normal DistributionsMean is 6 and standard deviation is 3