BUSN 352: Statistics Review

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## BUSN 352: Statistics Review

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**Professor Joseph Szmerekovsky**BUSN 352: Statistics Review Probability Distributions**Probability Concepts**• Let A be an event. Pr(A) is then the probability that A will occur… • If A never occurs, Pr(A) = 0 • If A is sure to occur, Pr(A) = 1**Example: Find the probability**• Selecting a black card from the standard deck of 52 cards • Selecting a King • Selecting a red King • General Pattern: Pr(A)= 1/2 4/52 =1/13 2/52 =1/26**Example: Find probability**Throwing two fair dice find the probability that • sum of the faces is equal to 2 • sum of the faces is equal to 4 • sum of the faces is equal to 7**Random Variables**• A random variable is a rule that assigns a numerical value to each possible outcome of an experiment. • Discrete random variable -- countable number of values. • Continuous random variable -- assumes values in intervals on the real line.**The Basics of Random Variables**• The probability distribution of a discrete random variable X gives the probability of each possible value of X. • The sum of the probabilities must be 1.**Example: Probability distributions**• Fair coin toss • Two fair coins • Find Pr(X1) • Find Pr(X=1.5) • Find Pr(X1.5)**Expected Value**• The expected value (mean) of the random variable is the sum of the products of x and the corresponding probabilities**Example : Insurance Policy**Alice sells Ben a $10,000 insurance policy at an annual premium of $460. If Pr(Ben dies next year) = .002, what is the expected profit of the policy? E(X) = 460(.998) + (-9540)(.002) = $440**Example: Debbon Air “Seat Release”**• Debbon Air needs to make a decision about Flight 206 to Myrtle Beach. • 3 seats reserved for last-minute customers (who pay $475 per seat), but the airline does not know if anyone will buy the seats. • If they release them now, they know they will be able to sell them all for $250 each.**Debbon Air “Seat Release”**• The decision must be made now, and any number of the three seats may be released. • Debbon Air counts a $150 loss of goodwill for every last-minute customer turned away. • Probability distribution for X = # of last-minute customers requesting seats:**“Debbon Air” Seat Release**• What is Debbon Air’s expected net revenue (revenue minus loss of goodwill) if all three seats are released now? • X = 0: Net Revenue = 3($250) = $750 • X = 1: Net Rev = 3($250) - $150 = $600 E (Net Revenue) = 750(.45) + 600(.30) + 450(.15) + 300(.10) = $615.**“Debbon Air” Seat Release**• How many seats should be released to maximize expected net revenue? Two seats should be released.**Variance and Standard Deviation of Random Variables**• The variance of a discrete R.V. X is • The standard deviation is the square root of the variance.**Probability density function**f(x) Area under the graph = Pr(a<X<b) a x b Continuous random variables • Continuous random variable -- assumes values in intervals on the real line. Total area = 1**f(x)**1 x 0.2 0.5 0.6 1 Example: Uniform distribution • Is this a valid probability density function? Yes • Find Pr(0.2 < X < 0.5) 0.3·1 = 0.3 • Find Pr(X > 0.6) 0.4·1 = 0.4**The Normal Probability Model**• Importance of the Normal model • Numerous phenomena seem to follow it, or can be approximated by it. • It provides the basis for classical statistical inference through the Central Limit Theorem. • It motivates the Empirical Rule.**The Normal Probability Model**• Crucial Properties • Bell-shaped, symmetric • Measures of central tendency (mean, median) are the same. • Parameters are mean and standard deviation .**The Normal Probability Model**The Normal probability density function: “The Bell Curve” fY(y) y**The Normal Probability Model**This area = 0.5 This area = fY(y) y a b**The Standard Normal Distribution**Normal with Mean SD Standard Normal with Mean 0 and SD 1 -2 -1 0 +1 +2**Table A.1: Standard Normal Distribution**• Standard Normal random variable Z • E(Z) = 0 and SD(Z) = 1 • Table A.1 gives Standard Normal probabilities to four decimal places. .4332 fZ(z) z 0 1.50**Practice with Table A.1**= .5 - .4332 = .0668 Pr(Z > 0) = .5 fZ(z) .4332 z 0 1.50**Practice with Table A.1**= .4332 + .5 = .9332 Pr(Z < 0) = .5 .4332 0 1.5**Practice with Table A.1**So k is about 1.645 .4500 .4495 k 0 1.64**Z Scores: Standardizing Normal Distributions**• Suppose X is • Transformation Formula: • For a given x, the Z score is the number of SD’s that x lies away from the mean.**Example: Tele-Evangelist Donations**• Money collected daily by a tele-evangelist, Y,is Normal with mean $2000, and SD $500. • What is the chance that tomorrow’s donations will be less than $1500? Convert to Z scores**Tele-Evangelist Donations**• Money collected is Normal with mean $2000 and SD $500. • What is the probability that tomorrow’s donations are between $2000 and $3000? • Let Y = $ collected tomorrow • Y is Normal with mean 2000 and SD 500 • Need : • Convert to Z scores: = .4772**Tele-Evangelist Donations**• What is the chance that tomorrow’s donations will exceed $3000? • Y is still Normal with mean 2000 and SD 500... Convert to Z scores