### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# INTRO BUSINESS STAT MTHSC 309

Clemson

GPA 3.76

### View Full Document

## 40

## 0

## Popular in Course

## Popular in Mathematics (M)

This 96 page Class Notes was uploaded by Jazmyn Braun on Saturday September 26, 2015. The Class Notes belongs to MTHSC 309 at Clemson University taught by Timothy Teitloff in Fall. Since its upload, it has received 40 views. For similar materials see /class/214284/mthsc-309-clemson-university in Mathematics (M) at Clemson University.

## Similar to MTHSC 309 at Clemson

## Reviews for INTRO BUSINESS STAT

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/26/15

Chapter 7 Random Variables and Discrete Probability Distributions Random Variables A random variable is a function or rule that assigns a number to each outcome of an experiment Alternatively the value of a random variable is a numerical event Instead of talking about the coin flipping event as heads tails think of it as the number of heads when flipping a coin 1 O numerical events Two Types of Random Variables Discrete Random Variable one that takes on a countable number of values Eg values on the roll of dice 2 3 4 12 Continuous Random Variable one whose values are not discrete not countable Eg time 301 minutes 3010000001 minutes Analogy Probability Distributions A probability distribution i a table formula or graph that describes the andom variable and the probability associated with these values Since we re describing a random variable which can be discrete or continuous we have two types of probability distributions Discrete Probability Distribution this chapter and Continuous Probability Distribution Chapter 8 Probability Notation An uppercase letter will represent the name of the random variable usually X Its lowercase counterpart will represent the value of the random variable The probability that the random variable X will equal X is PX X or more simply PX Discrete Probability Distributions The probabilities of the values of a discrete random variable may be derived by means of probability tools such as tree diagrams or by applying one ofthe definitions of probability so long as these tvvo conditions apply 10sPXslf0rallx 2213004 all x Example 71 The Statistical Abstract of the United States is published annually It contains a wide variety of information based on the census as well as other sources The objective is to provide information about a variety of different aspects of the lives of the country s residents One of the questions asked households to report the number of color televisions in the household The following table summarizes the data Develop the probability distribution of the random variable defined as the number of color televisions per household Example 71 Number of Color Televisions Number of Households 10008 1218 32379 37961 19387 7714 2842 Total 101501 Tlwa40 Example 71 Probability distributions can be estimated from relative frequencies eg PX4 P4 0076 6 Example 71 Eg what is the probability there is at least one television but no more than three in any given househnlr of It of at least one television but no more than three P1sXs 3 P1 P2 P3 319 374 191 884 Example 72 A mutual fund salesperson has arranged to call on three people tomorrow Based on past experience the salesperson knows that there is a 20 chance of closing a sale on each call Determine the probability distribution of the number of sales the salesperson will make Let 8 denote success ie closing a sale PS20 Thus SC is not closing a sale and PSC8O Example 72 Developing a Probability Distribution Sales Call 1 Sales Call 2 Sales Call 3 atmtmo32 8 X F x 3 23 008 3032096 1 3128384 o 83 512 PX2 is illustrated here PopulationProbability Distribution The discrete probability distribution represents a population Example 71 the population of number of TVs per household Example 72 the population of sales call outcomes Since we have populations we can describe them by computing various parameters Eg the population mean and population variance Population Mean Expected Value The population mean is the weighted average of all of its values The weights are the probabilities This parameter is also called the expected value ofX and is represented by EX EX u 2mm all I Population Variance The population variance is calculated similarly It is the weighted average of the squared deviations from the mean VX U2 2L1 MJEPLt if r I A 1 139 I11 AS before q WX UL 21 if all 1quot a v The standard deviation is the same as before Example 73 Find the mean variance and standard deviation for the population of the number of color televisions per household from Exarr 39 O f of Televisions Households in Px 0 1218 0 0012 1 32379 1 0319 2 37961 2 0374 3 19387 3 0191 4 7714 4 0076 5 2842 5 0028 101501 1000 EX p Expo 0 100 1c P1 5 135 0o121319 2374 3191 4076 5028 2084 Example 73 Find the mean variance and standard deviation for the population of the number of color televisions per household from Exarr 39 O f of Televisions Households 3 P00 0 1218 0 0012 1 32379 1 031 3l 2 37961 2 0374 3 19387 3 0191 4 7714 4 0076 5 2842 5 0028 101501 1000 VliX U2 Z x itfPLx HHJ 0 20842012 1 208423195 20842028 1107 Example 73 Find the mean variance and standard deviation for the population of the number of color televisions per household from Exarr 39 O f of Televisions Households in Px 0 1218 0 0012 1 32379 1 0319 2 37961 2 0374 3 19387 3 0191 4 7714 4 0076 5 2842 5 0028 101501 1000 039 quottier2 ll11107 1052 Laws of Expected Value Ec c The expected value of a constant c is just the value of the constant EX C EX C ECX CEX We can pull a constant out of the expected value expression either as part of a sum with a random variable X or as a coefficient of random variable X Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 1 Describe the problem statement in algebraic terms sales have a mean of 25000 ESales 25000 profits are calculated by Profit 30Sales 6000 Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of EProfit E30Sales 6000 E30Sales 6000 by rule 2 30ESales 6000 by rule 3 3025000 6000 1500 Thus the mean monthly profit is 1500 Laws of Variance Vc O The variance of a constant c is zero vmmwm The variance of a random variable and a constant is just the variance of the random variable per 1 above VcX C2VX The variance of a random variable and a constant coefficient is the coefficient squared times the variance of the random variable Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 1 Describe the problem statement in algebraic terms sales have a standard deviation of 4000 VSales 40002 16000000 rememta Relationship between standard deviation and variance profits are calculated by Profit 30Sales 6000 Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 6000 2 The variance of profit is VProfit V30Sales 6000 V30Sales by rule 2 302VSales by rule 3 30216000000 1440000 Again standard deviation is the square root of variance so standard deviation of Profit 1 440000 2 Example 74 summary Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 6000 Find the mean and standard deviation of monthly profits The mean monthly profit is 1500 The standard deviation of monthly profit is 1200 Bivariate Distributions Up to now we have looked at univariate distributions ie probability distributions in one variable As you might guess bivariate distributions are probabilities of combinations of two variables Bivariate probability distributions are also called joint probability A joint probability distribution ofX and Y is a table or formula that lists the joint probabilities for all pairs of values x and y and is denoted Pxy Pxy PXx and Yy Discrete Bivariate Distribution As you might expect the requirements for a bivariate distribution are similar to a univariate distribution with only minor changes to the notation 1 05PXysl 22 2Pxy1 all x all y for all pairs Xy Example 75 Xavier and Yvette are real estate agents Let X denote the number of houses that Xavier will sell in a month and let Y denote the number of houses Yvette will sell in a month An analysis of their past monthly performances has the following joint probabilities bivariate probability distribution H 0 1 E 0 012 042 000 06 15quot 1 021 006 003 03 2 00 002 001 01 04 05 01 100 Marginal Probabilities As before we can calculate the marginal probabilities by summing across rows and down columns to determine the probabilities of X and Y ihdividuquotquotquot x x O 1 2 molt o a g the probability that Xavier sells 1 house PX1 050 Describing the Bivariate Distribution We can describe the mean variance and standard deviation of each variable in a bivariate distribution by working with the marginal probabilities x Pltxgt v PW o o 4 0 o 5 1 O 5 1 o 3 2 O 1 2 o 1 E00 07 EV 5 vgtlt 041 VV 045 ox 064 067 same formulae as for univariate distributions Covariance The covariance of two discrete variables is defined as comm 2 Emmy MyPltXy quotHi all A or alternatively using this shortcut method COVXY 2 ExyPltxvgt My all A all v Coef cient of Correlation The coefficient of correlation is calculated in the same way as described earlier COVX Y 390 0X0 Example 76 Compute the covariance and the coefficient of correlation between the numbers of houses sold by COVXY 2 2m NM illiqpuuy UNI a 39139 COVXY o 70 512 1 70 542 2 72 501 15 E COVXY Uva There is a weak negative relationship between the two variables o1 5 6467 35 Example 76 SIMMM x x xooolx 1 o 1 2 o 1 2 o 1 g 05 Probabilitv X ux Y uv 012 07 05 021 07 05 007 07 15 042 03 05 006 03 05 002 03 15 006 13 05 003 13 05 001 13 15 X ugmeugyn 0042 0074 0074 0063 0009 0009 0039 0020 M 0150 Example 76 C0VX Y 0150 6467 OX0 35 There is a weak negative relationship between the two variables Sum of Two Variables The bivariate distribution allows us to develop the probability distribution of any combination of the two variables of particular interest is the sum of two variables If we consider our example of Xavier and Yvette selling houspc Aln nan nrnafn a nrnhahilihl Hicfrihllfinn x y 0 1 2 3 4 Pxy 012 063 019 005 001 to answer questions like what is the probability that two houses are sold PXY2 PO2 P11 P20 07 06 06 10 Sum of Two Variables Likewise we can compute the expected value variance and standard deviation of XY in the usual way EX Y O12163 219 305 4o1 12 VX Y o 12212 4 122o1 56 cm 1VarXY J 75 Laws We can derive laws of expected value and variance for the sum of two variables as follows EX Y EX EY VX Y VX VY ZCOVX Y IfX and Y are independent COVX Y O and thus VX Y VX VY Laws EXY EX EY 7 5 12 VX Y VX VY ZCOVX Y 41 45 215 56 Portfolio Diversification and Asset Allocation Consider an investor who forms a portfolio consisting of only two stocks by investing 4000 in one stock and 6000 in a second stock Suppose that the results after 1 year are OneYear Results Initial Value of Investment Rate of Return Stock Investment After One Year on Investment 1 4000 5000 R1 25 25 2 6000 5400 R2 10 10 Total 10000 10400 RIo 04 4 OP Portfolio Diversification and Asset Allocation Mean and Variance of a Portfolio of Two Stocks ERp W1 ER1 W2 ER2 VRp W12 VR1 W22 VR2 2W1W2 COVR1 R2 W12012 W22022 2W1W2F30102 where w1 and w2 are the proportions or weights of investments 1 and 2 ER1 and ER2 are their expected values 01 and 02 are their standard deviations and p is the coefficient of correlation Example 78 An investor has decided to form a portfolio by putting 25 of his money into McDonald s stock and 75 into Cisco Systems stock The investor assumes that the expected returns will be 8 and 15 respectively and that the standard deviations will be 12 and 22 respectively a Find the expected return on the portfolio b Compute the standard deviation of the returns on the portfolio assuming that i the two stocks returns are perfectly positively correlated ii the coefficient of correlation is 5 iii the two stocks returns are uncorrelated Example 78 Solution a The expected values of the two stocks are ER1 08 and ER2 15 The weights are w1 25 and w2 75 Thus ER2 W1ER1 W2ER2 25c0875c15 1325 Example 78 Solution The standard deviations are 01 12 and 02 22 Thus VRp W12012 W22022 2W1W2F30102 252122 752222 22575p 1222 0281 0099 p When p 1 VRp 0281 00991 0380 When p 5 VRp 0281 00995 0331 When p 0 VRp 0281 00990 0281 Portfolio Diversification in Practice The formulas introduced in this section require that we know the expected values variances and covariance or coefficient of correlation of the investments we re interested in The question arises How do we determine these parameters Incidentally this question is rarely addressed in finance textbooks The most common procedure is to estimate the parameters from historical data using sample statistics Portfolios with More Than Two Stocks We can extend the formulas that describe the mean and variance of the returns of a portfolio of two stocks to a portfolio of any number of stocks Mean and Variance of a Portfolio of k Stocks EltRp EvaEat VRp iwfcf 2Zk WiWJCOVRiRJ i1 i1 ji1 Where RJ is the return of the ith stock wi is the proportion of the portfolio Invested in stock i and k is the number of stocks in the port olio Portfolios with More Than Two Stocks When k is greater than 2 the calculations can be tedious and timeconsuming For example when k 3 we need to know the values of the three weights three expected values three variances and three covariances When k 4 there are four expected values four variances and six covariances The number of covariances required in general is kk12 To assist you we have created an Excel worksheet to perform the computations when k 2 3 or 4 For larger values of k see the reference at the end of the chapter To demonstrate we ll return to the problem described in this chapter s introduction ChapterOpening Example An investor has 100000 to invest in the stock market She is interested in developing a stock portfolio made up of General Electric General Motors McDonald s and Motorola However she doesn t know how much to invest in each one She wants to maximize her return but she would also like to minimize the risk She has computed the monthly returns for all four stocks during a 60month period January 2001 to December 2006 Xm0700 ChapterOpening Example After some consideration she narrowed her choices down to the following three What should she do 1 25000 in each stock 2 General Electric 10000 General Motors 20000 McDonald s 30000 Motorola 40000 3 General Electric 10000 General Motors 50000 McDonald s 30000 Motorola 10000 ChapterOpening Example Because of the large amount of calculations we will solve this problem using only Excel From the file we compute the means of each stock s returns B C D E 74 0000305 0002339 0007910 0007997 ChapterOpening Example Next we compute the variancecovariance matrix The commands are the same as those described in Chapter 4 simply include all the columns of the returns of the investments you wish to include in the portfolio A B C D E 1 GE GM McDonalds Motorola 2 GE 0003493 3 GM 0001076 0011016 4 McDonalds 0001528 0001989 0005409 5 Motorola 0000933 0004131 0002515 0010277 ChapterOpening Example Notice that the variances of the returns are listed on the diagonal Thus for example the variance of the 60 monthly returns of General Electric is 00349 The covariances appear below the diagonal The covariance between the returns of General Electric and General Motors is 00108 The means and the variancecovariance matrix are copied to the Portfolio Diversification spreadsheet The weights are typed producing the accompanying output ChapterOpening Example A B C D E F 1 Portfolio of4Stocks 2 GE GM McDonalds Motorola 3 VarianceCovariance Matrix GE 0003493 4 GM 0001076 0011016 5 McDonalds 0001528 0001989 0005409 6 Motorola 0000933 0004131 0002515 0010277 7 8 Expected Returns 0000305 0002339 0007910 0007997 9 10 Weights 0250000 0250000 0250000 0250000 11 12 Portfolio Return 13 Expected Value 00046 14 Variance 00034 15 Standard Deviation 00584 ChapterOpening Example Plan 2 A B C D E F 1 Portfolio of4Stocks 2 GE GM McDonalds Motorola 3 VarianceCovariance Matrix GE 0003493 4 GM 0001076 0011016 5 McDonalds 0001528 0001989 0005409 6 Motorola 0000933 0004131 0002515 0010277 7 8 Expected Returns 0000305 0002339 0007910 0007997 9 10 Weights 0100000 0200000 0300000 0400000 11 12 Portfolio Return 13 Expected Value 00061 14 Variance 00043 15 Standard Deviation 00657 ChapterOpening Example Plan 3 A B C D E F 1 Portfolioof4Stocks 2 GE GM McDonalds Motorola 3 VarianceCovariance Matrix GE 0003493 4 GM 0001076 0011016 5 McDonalds 0001528 0001989 0005409 6 Motorola 0000933 0004131 0002515 0010277 7 8 Expected Returns 0000305 0002339 0007910 0007997 9 10 Weights 0100000 0500000 0300000 0100000 11 12 Portfolio Return 13 Expected Value 00044 14 Variance 00048 15 Standard Deviation 00690 ChapterOpening Example Plan 3 has the smallest expected value and the largest variance making it the worst of the three plans Plan 2 has the largest expected value whereas plan 1 has the smallest variance lfthe investor is like most investors she would select Plan 1 because of its lower risk Other more daring investors may choose plan 2 to take advantage of its higher expected value Chapter 7 Random Variables and Discrete Probability Distributions Random Variables A random variable is a function or rule that assigns a number to each outcome of an experiment Alternatively the value of a random variable is a numerical event Instead of talking about the coin flipping event as heads tails think of it as the number of heads when flipping a coin 1 O numerical events Two Types of Random Variables Discrete Random Variable one that takes on a countable number of values Eg values on the roll of dice 2 3 4 12 Continuous Random Variable one whose values are not discrete not countable Eg time 301 minutes 3010000001 minutes Analogy Probability Distributions A probability distribution is a table formula or graph that describes the val 6 variable and the probability associated with these values Since we re describing a random variable which can be discrete or continuous we have two types of probability distributions Discrete Probability Distribution this chapter and Continuous Probability Distribution Chapter 8 Probability Notation An uppercase letter will represent the name of the random variable usually X Its lowercase counterpart will represent the value of the random variable The probability that the random variable X will equal X is PX X or more simply PX Discrete Probability Distributions The probabilities of the values of a discrete random variable may be derived by means of probability tools such as tree diagrams or by applying one ofthe definitions of probability so long as these tvvo conditions apply 10sPXslf0rallx 2213004 all x Example 71 The Statistical Abstract of the United States is published annually It contains a wide variety of information based on the census as well as other sources The objective is to provide information about a variety of different aspects of the lives of the country s residents One of the questions asked households to report the number of color televisions in the household The following table summarizes the data Develop the probability distribution of the random variable defined as the number of color televisions per household Example 71 Number of Color Televisions Number of Households 10008 1218 32379 37961 19387 7714 2842 Total 101501 Tlwa40 Example 71 Probability distributions can be estimated from relative frequencies eg PX4 P4 0076 6 Example 71 Eg what is the probability there is at least one television but no more than three in any given househnlr of It of at least one television but no more than three P1sXs 3 P1 P2 P3 319 374 191 884 PopulationProbability Distribution The discrete probability distribution represents a population Example 71 the population of number of TVs per household Example 72 the population of sales call outcomes Since we have populations we can describe them by computing various parameters Eg the population mean and population variance Expected Value The expected value of a random variable is the weighted average of all of its values The weights are the probabilities EX p 211305 allx Interpretation If the experiment upon which X is based is repeated a large numbers of times and the computed average of these trials is expected to be close to EX Variance The variance of a random variable X is the weighted average of the squared deviations from the mean WX a 20 ri39fPt x HHI There is a shortcut formula WK cri EngEma if x The standard deviation is o Example 73 Find the mean variance and standard deviation for the population of the number of color televisions per household from Exarr 39 O f of Televisions Households in Px 0 1218 0 0012 1 32379 1 0319 2 37961 2 0374 3 19387 3 0191 4 7714 4 0076 5 2842 5 0028 101501 1000 EX p Expo 0 100 1c P1 5 135 0o121319 2374 3191 4076 5028 2084 Example 73 Find the mean variance and standard deviation for the population of the number of color televisions per household from Exarr 39 O f of Televisions Households 3 P00 0 1218 0 0012 1 32379 1 031 3l 2 37961 2 0374 3 19387 3 0191 4 7714 4 0076 5 2842 5 0028 101501 1000 VliX U2 Z x itfPLx HHJ 0 20842012 1 208423195 20842028 1107 Example 73 Find the mean variance and standard deviation for the population of the number of color televisions per household from Exarr 39 O f of Televisions Households in Px 0 1218 0 0012 1 32379 1 0319 2 37961 2 0374 3 19387 3 0191 4 7714 4 0076 5 2842 5 0028 101501 1000 039 quottier2 ll11107 1052 Laws of Expected Value Ec c The expected value of a constant c is just the value of the constant EX C EX C ECX CEX We can pull a constant out of the expected value expression either as part of a sum with a random variable X or as a coefficient of random variable X Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 1 Describe the problem statement in algebraic terms sales have a mean of 25000 ESales 25000 profits are calculated by Profit 30Sales 6000 Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of EProfit E30Sales 6000 E30Sales 6000 by rule 2 30ESales 6000 by rule 3 3025000 6000 1500 Thus the mean monthly profit is 1500 Laws of Variance Vc O The variance of a constant c is zero vmmwm The variance of a random variable and a constant is just the variance of the random variable per 1 above VcX C2VX The variance of a random variable and a constant coefficient is the coefficient squared times the variance of the random variable Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 1 Describe the problem statement in algebraic terms sales have a standard deviation of 4000 VSales 40002 16000000 rememta Relationship between standard deviation and variance profits are calculated by Profit 30Sales 6000 Example 74 Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 6000 2 The variance of profit is VProfit V30Sales 6000 V30Sales by rule 2 302VSales by rule 3 30216000000 1440000 Again standard deviation is the square root of variance so standard deviation of Profit 1 440000 2 Example 74 summary Monthly sales have a mean of 25000 and a standard deviation of 4000 Profits are calculated by multiplying sales by 30 and subtracting fixed costs of 6000 Find the mean and standard deviation of monthly profits The mean monthly profit is 1500 The standard deviation of monthly profit is 1200 Binomial Distribution The binomial distribution is the probability distribution that results from doing a binomial experiment 1 Fixed number of trials represented as n 2 Each trial has two possible outcomes one called the success and the other the failure 3 The probability of a success is the same for each trial Psuccess p and Pfailure 1 p q 4 The trials are independent which means that the outcome of one trial does not affect the outcomes of any other trials Binomial Random Variable The binomial random variable is discrete The probability distribution for a binomial rv is defined with a formula 1 quot X Px p 117 forx 0 1 2 n xln xl n is called nfactorial It represents the product of all whole numbers less than or equal to n nnn 1n 2321 Pat Statsdud Pat Statsdud is a not good student taking a statistics course Pat s exam strategy is to rely on luck for the next quiz The quiz consists of 10 multiplechoice questions Each question has five possible answers only one of which is correct Pat plans to guess the a gotinn A A A m w c ab What is the probability that Pat gets no answers correct What is the probability that Pat gets two answers correct Pat Statsdud Pat Statsdud is a not good student taking a statistics course whose exam strategy is to rely on luck for the next quiz The quiz consists of 10 multiplechoice questions Each question has five possible answers only one of which is correct Pat plans to guess the answeW eh quecticn Algebraically then n10 and Psuccess 15 20 Pat Statsdud Is this a binomial experiment Check the conditions There is a fixed finite number of trials n10 An answer can be either correct or incorrect The probability of a correct answer Psuccess20 does not change from question to question Each answer is independent of the others Pat Statsdud n10 and Psuccess 20 What is the probability that Pat gets no answers correct e SL 10 0 10 ON Px0 P 0 2 1 2 01004 11810 1074 Pat has about an 11 chance of getting no answers correct using the guessing strategy Pat Statsdud n10 and Psuccess 20 What is the probability that Pat gets two answers correct e success x 2 hence we want to know Px2 P2 10 221 2W 2110 2 39 39 45041678 3020 Pat has about a 30 chance of getting exactly two answers correct using the guessing strategy Cumulative Probability Thus far we have been using the binomial probability distribution to find probabilities for individual values of x To answer the question Find the probability that Pat fails the qu39 requires a cumulative probability that is PX s x If a grade on the quiz is less than 50 Le 5 questi ns out of 10 that s considered a failed quiz Thus we want to know what is PX S 4 to answer Pat Statsdud PX s 4 P0 P1 P2 P3 P4 We already know P0 1074 and P2 3020 Using the binomial formula to calculate the others P1 2684 P3 2013 and P4 0881 We have PX S 4 1074 2684 0881 9672 Thus its about 97 probable that Pat will fail the test using the luck strategy and guessing at Table 1 in Appendix B list cummulative probabilities for several values of n p and kfor binomial rvs Example What is PX S 4 given Psuccess 20 and n10 PX s 4 09672 Binomial Table The probabilities listed in the tables are cumulative ie PX s k k is the row index the columns of the table are organized by Psuccess p 001 005 01 09044 05987 03487 09957 09139 0736 09999 09885 092 10000 09990 098 0000 09999 099 0000 10000 099 025 03 04 05 06 07 075 08 09 095 099 00563 00282 00060 00010 00001 00000 00000 00000 00000 00000 00000 02440 01493 00464 00107 00017 00001 00000 00000 00000 00000 00000 5256 03828 01673 00547 00123 00016 00004 00001 00000 00000 00000 7759 06496 03823 01719 00548 00106 00035 00009 00000 00000 00000 9219 08497 06331 03770 01662 00473 00197 00064 00001 00000 00000 9803 09527 08338 06230 03669 01503 00781 00328 00016 00001 00000 0000 10000 1000 09965 09894 09452 08281 06177 03504 02241 01209 00128 00010 00000 0000 10000 1000 09996 09984 09877 09453 08327 06172 04744 03222 00702 00115 00001 0000 10000 10000 10000 09999 09983 09893 09536 08507 07560 06242 02639 00861 00043 0000 10000 10000 0000 10000 09999 09990 09940 09718 09437 08926 06513 04013 00956 Binomial Table What is the probability that Pat gets no answers correct ie what is PX 0 given Psuccess 025 03 04100539 06 07 075 08 n10 001 095 01 02 09044 05987 0348quot 01074 k 0 1 2 3 4 5 6 7 8 9 09957 09999 10000 09139 09885 09990 09999 10000 07361 09293 09872 09984 09999 1 0000 1 0000 1 0000 1 0000 03758 06778 08791 09672 09936 09991 09999 10000 10000 PX 0 PX s 0 1074 00563 02440 05256 07759 09219 09803 09965 09996 10000 10000 00282 01493 03828 06496 08497 09527 09894 09984 09999 10000 00060 00464 01673 03823 06331 08338 09452 09877 09983 09999 00010 00107 00547 01719 03770 06230 08281 09453 09893 09990 00001 00017 00123 00548 01662 03669 06177 08327 09536 09940 00000 00001 00016 00106 00473 01503 03504 06172 08507 09718 00000 00000 00004 00035 00197 00781 02241 04744 07560 09437 00000 00000 00001 00009 00064 00328 01209 03222 06242 08926 09 00000 00000 00000 00000 00001 00016 00128 00702 02639 06513 095 00000 00000 00000 00000 00000 00001 00010 00115 00861 04013 20 and 099 00000 00000 00000 00000 00000 00000 00000 00001 00043 00956 Binomial Table What is the probability that Pat gets two answers orrect ie what is P X 2 given Psuccess 20 and n10 k OWNGgtUIJgtWNA 00282 01498 08828 06496 08497 09527 09894 09984 09999 10000 PX 2 P remember the table 00060 00464 01678 08828 06881 08888 09452 09877 09988 09999 39 ll is 7 00010 00107 00547 01719 08770 06280 08281 09458 09898 09990 00001 00017 00128 00548 01662 08669 06177 08827 09586 09940 00000 00001 00016 00106 00478 01508 08504 06172 08507 09718 00000 00000 00004 00085 00197 00781 02241 04744 07560 09487 00000 00000 00001 00009 00064 00828 01209 08222 06242 08926 1 6778 3758 shows umuative probabilities 00000 00000 00000 00000 00001 00016 00128 00702 02689 06518 00000 00000 00000 00000 00000 00001 00010 00115 00861 04018 3020 025 3905 06 07 075 08 09 095 099 00000 00000 00000 00000 00000 00000 00000 00001 00048 00956 Binomial Table The binomial table gives cumulative probabilities for PXs k but as we ve seen in the last example Pk PX s k PX s k 1 Likewise for probabilities given as PX 2 k we have Pk 1 PX s k 1 BINOMDIST Excel Function There is a binomial distribution function in Excel that can also be used to calculate these probabilities For example Vll x v MaulsHEJOPZJALSET etS SUCCESSGS BlNOMDIST NumberJ 2 Trials m E 13 lt ProhahilityJ 2 E 32 Cumulative FALSE E FALSE Psuccess Returns the individual term binomial distribution probability cu m ulatlve Cumulative is a logical value for the cumulative distribution function use e TRUE for the probability mass function use FALSE 39 39 39 Formula result 3331333333 Cancel El PX23020 BINOMDIST Excel Function There is a binomial distribution function in Excel that can also be used to calculate these probabilities For example thalamus the quiz successes m El z m lt trials PruhlhllllyA z E U2 mm Irma El m Psuccess cumulatlve i alumni human lm in ms umuhuvz isnl mm funk mu arm mummy mm mnman swim usmzussuz PXS49672 Binomial Distribution As you might expect statisticians have developed general formulas for the mean variance and standard deviation of a binomial random variable They are l P 02 np p Oquot qliip p

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.