Statistics for Business and Economics Chapter 4 Random Variables & Probability Distributions

Learning Objectives Distinguish Between the Two Types of Random Variables Describe Discrete Probability Distributions Describe the Uniform and Normal Distributions As a result of this class, you will be able to...

Learning Objectives (continued) Explain Sampling Distributions Solve Probability Problems Involving Sampling Distributions

Types of Random Variables

Data Types Data Quantitative Qualitative Continuous Discrete

Discrete Random Variables

Data Types Data Quantitative Qualitative Continuous Discrete

Discrete Random Variable A numerical outcome of an experiment Example: Number of tails in 2 coin tosses Discrete random variable Whole number (0, 1, 2, 3, etc.) Obtained by counting Usually a finite number of values Poisson random variable is exception ()

Discrete Random Variable Examples Possible Values Experiment Make 100 Sales Calls # Sales 0, 1, 2, ..., 100 Inspect 70 Radios # Defective 0, 1, 2, ..., 70 Answer 33 Questions # Correct 0, 1, 2, ..., 33 Count Cars at Toll Between 11:00 & 1:00 # Cars Arriving 0, 1, 2, ..., ∞

Continuous Random Variables

Data Types Data Quantitative Qualitative Continuous Discrete

Continuous Random Variable A numerical outcome of an experiment Weight of a student (e.g., 115, 156.8, etc.) Continuous Random Variable Whole or fractional number Obtained by measuring Infinite number of values in interval Too many to list like a discrete random variable

Continuous Random Variable Examples Possible Values Experiment Weigh 100 People Weight 45.1, 78, ... Measure Part Life Hours 900, 875.9, ... Amount spent on food $ amount 54.12, 42, ... Measure Time Between Arrivals Inter-Arrival Time 0, 1.3, 2.78, ...

Probability Distributions for Discrete Random Variables

Discrete Probability Distribution List of all possible [x, p(x)] pairs x = value of random variable (outcome) p(x) = probability associated with value Mutually exclusive (no overlap) Collectively exhaustive (nothing left out) 0  p(x)  1 for all x  p(x) = 1

Discrete Probability Distribution Example Experiment: Toss 2 coins. Count number of tails. Probability Distribution Values, x Probabilities, p(x) 0 1/4 = .25 1 2/4 = .50 2 1/4 = .25 © 1984-1994 T/Maker Co.

Visualizing Discrete Probability Distributions Listing Table { (0, .25), (1, .50), (2, .25) } f(x) p(x) # Tails Count 1 .25 1 2 .50 Experiment is tossing 1 coin twice. Graph 2 1 .25 p(x) .50 Formula .25 x n ! .00 p ( x ) = px(1 – p)n - x 1 2 x!(n – x)!

Summary Measures Expected Value (Mean of probability distribution) Weighted average of all possible values  = E(x) = x p(x) Variance Weighted average of squared deviation about mean 2 = E[(x (x p(x) population notation is used since all values are specified. 3. Standard Deviation ●

Summary Measures Calculation Table x p(x) x p(x) x –  (x – )2 (x – )2p(x) Total xp(x) (x p(x)

Thinking Challenge You toss 2 coins. You’re interested in the number of tails. What are the expected value, variance, and standard deviation of this random variable, number of tails? © 1984-1994 T/Maker Co.

Expected Value & Variance Solution* p(x) x p(x) x –  (x – ) 2 (x – ) 2p(x) .25 .50  = 1.0 -1.00 1.00 .25 2 = .50  = .71 1 .50 2 .25 1.00 1.00

Portfolio Selection Case

Portfolio Selection Case

Data Types Data Quantitative Qualitative Continuous Discrete

Probability Distributions for Continuous Random Variables

Continuous Probability Density Function Mathematical formula Shows all values, x, and frequencies, f(x) f(x) Is Not Probability Value (Value, Frequency) Frequency f(x) a b x (Area Under Curve) f x dx ( ) All x a b     1 0, Properties

Continuous Random Variable Probability ( a  x  b )   f ( x ) dx Probability Is Area Under Curve! a f(x) x a b © 1984-1994 T/Maker Co.

Continuous Probability Distributions Uniform Normal

Uniform Distribution x f(x) d c a b 1. Equally likely outcomes 2. Probability density function 3. Mean and Standard Deviation

Uniform Distribution Example You’re production manager of a soft drink bottling company. You believe that when a machine is set to dispense 12 oz., it really dispenses 11.5 to 12.5 oz. inclusive. Suppose the amount dispensed has a uniform distribution. What is the probability that less than 11.8 oz. is dispensed? SODA

Uniform Distribution Solution f(x) 1.0 x 11.5 11.8 12.5 P(11.5  x  11.8) = (Base)(Height) = (11.8 - 11.5)(1) = .30

Normal Distribution

Importance of Normal Distribution Describes many random processes or continuous phenomena Can be used to approximate discrete probability distributions Example: binomial Basis for classical statistical inference

Normal Distribution f ( x ) x ‘Bell-shaped’ & symmetrical Mean, median, mode are equal IQR is 1.33 Random variable has infinite range f ( x ) x Mean Median Mode

Probability Density Function f(x) = Frequency of random variable x  = Population standard deviation  = 3.14159; e = 2.71828 x = Value of random variable (– < x < )  = Population mean

Effect of Varying Parameters ( & ) f(X) B A C X

Normal Distribution Probability Probability is area under curve! f ( x ) Use Minitab!!! x c d

Using Minitab for Normal Probabilities Minitab only gives the cumulative probability P(X < UB) where “UB” is an upper bound (and LB is lower bound). UB Three hints: P(X < UB) (Minitab gives this directly) P(LB < X < UB) = P(X < UB) - P(X < LB) (So use Minitab twice and compute the difference P(X > LB) = 1 - P(X < UB) (So use Mintab once and then subtract from 1.0 using the complimentation law)

Using Minitab for Normal Probabilities X follows N(100,) distribution; find P(X < 115) Minitab menu sequence: CALC >> Probability Distributions >> Normal m=100 and s2=25 so s=5

Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000 hours and = 200 hours. What’s the probability that a bulb will last: A. Between 2100 and 2400 hours? Less than 1470 hours? More than 2500 hours Greater than 2000 hours Allow students about 10-15 minutes to solve this.

Finding Normal Percentiles The .35 percentile: Find the X-value such that 35% of the population falls below this value and 65% fall above it. .35 .65 X This is just the INVERSE of finding normal probabilities: Before: Given an X value, find the cumulative probability Now: Given a cumulative probability, find the X value In Minitab, we use the “Inverse cumulative probability” option

Problem: Find the .35 percentile for a normal distribution with mean 20 and standard deviation 5. Need to click the “Inverse” button!! Then put the percentile you need here.

Reliability Example Life testing has revealed that a particular type of TV picture tube has a length of life that is approximately normally distributed with a mean of 8000 hours and a standard deviation of 1000 hours. The manufacturer wants to set a guarantee period for the tube that will obligate the manufacturer to replace no more than 5% of all tubes sold. How long should the guarantee period be?

Assessing Normality

Assessing Normality Draw a histogram or stem–and–leaf display and note the shape Compute the intervals x + s, x + 2s, x + 3s and compare the percentage of data in these intervals to the Empirical Rule (68%, 95%, 99.7%) Calculate If ratio is close to 1.3, data is approximately normal

Assessing Normality Continued Draw a Normal Probability Plot Observed value Expected Z–score

Checking for Normality Construct a “normal probability plot” of the data. If the data are approximately normal, the points will fall approximately on a straight line. Suppose the sample has mean X and standard deviation s. Then the normal probability plot plots: X Axis: Actual value (and suppose its percentile is p) Y Axis: The pth percentile from an actual normal distribution with mean X and standard deviation s (i.e., the “expected normal value”) _ _

Checking for Normality This is sometimes called a normal quantile-quantile plot

Checking for Normality An absolutely equivalent plot can be obtained by re-labeling the Y axis using the percentiles rather than the normal values corresponding to the percentiles. This gives a normal probability plot and is what Minitab does: Different Label: Same Plot! If “P-value” is less than .05, conclude NOT normal

Normal Plots of Residuals N = 32 Normally Distributed Residuals Bell Shape Straight Line

Normal Plots of Residuals: Patterns Outliers on both sides: “S” Shape Investigate outliers Skewed right: curving down Take log of Y (quite common) Skewed left: curving up Rarely happens

Sampling Distributions

Parameter & Statistic Parameter Sample Statistic Summary measure about population Sample Statistic Summary measure about sample P in Population & Parameter S in Sample & Statistic

Common Statistics & Parameters Sample Statistic Population Parameter Mean  X  Standard Deviation s  Variance s2 2 Binomial Proportion p ^

Sampling Distribution Theoretical probability distribution Random variable is sample statistic Sample mean, sample proportion, etc. Results from drawing all possible samples of a fixed size 4. List of all possible [x, p(x)] pairs Sampling distribution of the sample mean

Developing Sampling Distributions Suppose There’s a Population ... Population size, N = 4 Random variable, x Values of x: 1, 2, 3, 4 Uniform distribution © 1984-1994 T/Maker Co.

Population Characteristics Summary Measures Population Distribution P(x) .3 .2 Have students verify these numbers. .1 .0 x 1 2 3 4

All Possible Samples of Size n = 2 2nd Observation 1 2 3 4 1st Obs 16 Sample Means 2nd Observation 1 2 3 4 1st Obs 1,1 1,2 1,3 1,4 1.0 1.5 2.0 2.5 2,1 2,2 2,3 2,4 1.5 2.0 2.5 3.0 3,1 3,2 3,3 3,4 2.0 2.5 3.0 3.5 4,1 4,2 4,3 4,4 2.5 3.0 3.5 4.0 Sample with replacement

Sampling Distribution of All Sample Means 2nd Observation 1 2 3 4 1st Obs 16 Sample Means Sampling Distribution of the Sample Mean .0 .1 .2 .3 1.0 1.5 2.0 2.5 3.0 3.5 4.0 P(x) x 1.0 1.5 2.0 2.5 1.5 2.0 2.5 3.0 2.0 2.5 3.0 3.5 2.5 3.0 3.5 4.0

Summary Measures of All Sample Means Have students verify these numbers.

Sampling Distribution Comparison Population Sampling Distribution P(x) .0 .1 .2 .3 1 2 3 4 .0 .1 .2 .3 1.0 1.5 2.0 2.5 3.0 3.5 4.0 P(x) x x

Standard Error of the Mean 1. Standard deviation of all possible sample means, x ● Measures scatter in all sample means, x Less than population standard deviation 3. Formula (sampling with replacement)

Properties of the Sampling Distribution of x Regardless of the sample size, The mean of the sampling distribution equals the population mean An estimator is a random variable used to estimate a population parameter (characteristic). Unbiasedness An estimator is unbiased if the mean of its sampling distribution is equal to the population parameter. Efficiency The efficiency of an unbiased estimator is measured by the variance of its sampling distribution. If two estimators, with the same sample size, are both unbiased, then the one with the smaller variance has greater relative efficiency. Consistency An estimator is a consistent estimator of a population parameter if the larger the sample size, the more likely it is that the estimate will come close to the parameter. The standard deviation of the sampling distribution equals 3. And what about the shape of the sampling distribution?

Central Limit Theorem X As sample size gets large enough (n  30) ... sampling distribution becomes almost normal. X

Population (Probability Dist’n)

Sample Size = 2

Sample Size = 4

Sample Size = 8

Sample Size = 16

Sample Size = 32

Sample Size = 1 (Population)

Sample Size= 2

Sample Size= 8

Sample Size = 16

Sample Size = 32

Summary: Sampling from Normal or Non-Normal Populations Central Tendency Dispersion Sampling with replacement Population Distribution s = 10 m = 50 X Sampling Distribution n = 4 X = 5 n =30 X = 1.8 m - = 50 X X

Thinking Challenge You’re an operations analyst for AT&T. Long-distance telephone calls are normally distribution with  = 8 min. and  = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes? © 1984-1994 T/Maker Co.

Central Limit Theorem Example The amount of soda in cans of a particular brand has a mean of 12 oz and a standard deviation of .2 oz. If you select random samples of 50 cans, what percentage of the sample means would be less than 11.95 oz? SODA

Conclusion Distinguished Between the Two Types of Random Variables Described Discrete Probability Distributions Described the Uniform and Normal Distributions Explained Sampling Distributions Solved Probability Problems Involving Sampling Distributions As a result of this class, you will be able to...