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Review of Ch15 In Ch15, we learned General Addition Rule Conditional Probability General Multiplication Rule Redefine the Independence Tree diagram Next, we are going to start Ch16 Random Variables o o o Definition of a random variable Discrete random variables and continuous random variables Expectation, variance and standard deviation of the random variable Ch16 Random Variables Let’s look at the following example first Suppose you participate in a game of rolling a fair dice. If you get a six, then you will win $4; if you get a five, then you will win $2; if you get any other number, you will win nothing. The cost for this game is $1 per time. Then what’s your net income from this game? Number you Roll Net Income Probability 6 $4 - $1= $ 3 1/6 5 $2 - $1= $ 1 1/6 1,2,3,4 $0 - $1= - $1 4/6=2/3 Term 1: Random Variables A numerical value, which is based on the outcome of a random event, is called a random variable. e.g. The Net Income is a typical random variable. A random variable is usually denoted by a capital letter, like X, Y or Z and so on. We denote the values that a random variable can take by using the corresponding lower-case letter. For example, if you are going to denote the random variable “Net Income” by using X, then any particular value of the “Net Income” will be denoted by x Number you roll Net Income x Probability 6 $4 - $1= $ 3 1/6 5 $2 - $1= $ 1 1/6 1,2,3,4 $0 - $1= - $1 4/6=2/3 Term 1: Random Variables If we can list all the outcomes, we might formally call this random variable a discrete random variable. Otherwise, we’d call it a continuous random variable. e.g. The “Net Income” is a discrete random variable The “Height” is a continuous random variable The collection of all the possible values and the probabilities that they occur is called the probability model for the discrete random variable. The probability that a particular outcome x will happen is usually denoted by P(X=x) or P(x) This is the probability model for the random variable “Net Income” Net Income x Probability P(X=x) $3 1/6 $1 1/6 - $1 2/3 Term 1: Random Variables Let’s try an example together An insurance company offers a “death and disability” policy that pays $10,000 when someone dies or $5,000 if someone is permanently disabled. It charges a premium of $50 a year for this benefit. Based on a report, the death rate in any year is 1 out of every 1000 people, and that another 2 out of 1000 suffer some kind of disability. Question: Please figure out the probability model for the company’s net payout in a given year. Solution: Net Payout x Probability P(x) $10,000-$50=$9,950 1/1000=0.001 $5,000-$50=$4,950 2/1000=0.002 $0 - $50 = - $50 1-0.001-0.002=0.997 Term 1: Random Variables Let’s try another example together Suppose a random variable X has the following probability model: Q1: What are the all possible values of the x P(X=x) random variable X? 1 0.1 A1: 1,3,5,6,8,10 3 0.2 Q2: P(X=3)= ? P(X=2)=? P(X≤ 5) = ? 5 0.15 P(2≤X<8) =? P(X>7)=? 6 0.05 A2: P(X=3)=0.2 P(X=2)=0 8 0.3 P(X≤ 5) =0.1+0.2+0.15=0.45 10 0.2 P(2≤X<8) =0.2+0.15+0.05=0.4 P(X>7)=0.3+0.2=0.5 Term 2: Expectation The expectation of a random variable X is often denoted by E(X) or The formula for computing the expectation of a discrete random variable is given by that is to multiply each possible value by the probability that it occurs and find the sum. Sometimes, the expectation of a random variable is also called the mean of the random variable Expectation gives us the measure of the center for the values of a random variable w.r.t. the probability weight. Term 2: Expectation Let’s try an example together. With the probability model of a random variable X, please find: Q1: E(X) A1: Q2:E(X2) A2: x P(X=x) 1 0.1 3 0.2 5 0.15 6 0.05 8 0.3 10 0.2 Term 2: Expectation Let’s try one more example Please find the expectation of the random variable X of “Net Income” and also comment on the expectation. Net Income x Probability P(X=x) $3 1/6 $1 1/6 - $1 2/3 Solution: This tells you: Even though you can get some profit from the game occasionally, you still can’t get any profit in the long run. Term 3: Standard Deviation The variance of a random variable X is denoted by Var(X) or . And the variance of a discrete random variable given by Var ( X ) = σ 2 = ∑ ( x − µ ) 2 P ( x) = E ( X 2 ) − [ E ( X )]2 i.e. variance is the expected value of those squared deviations The standard deviation of a random variable X is denoted by SD(X) or . And the standard deviation of a discrete random variable is given by The standard deviation is a measure of the spread of the values of a random variable Var(X)=SD(X)2 or equivalently, Term 2: Expectation Let’s try one example Please find the variance and the standard deviation of the random variable X of “Net Income” Net Income x Probability P(X=x) $3 1/6 $1 1/6 - $1 2/3 Solution: This is the variance This is the standard deviation How to use TI to find E(X) and SD(X)? σx Net Income x Probability P(X=x) $3 1/6 $1 1/6 - $1 2/3 Please find the E(X) and SD(X) by using TI. More about E(X), Var(X) and SD(X) Shifting and Rescaling on a Random Variable Shifting Rescaling Shifting will make all the measures of center (like mean) to be shifted, but keep all the measures of spread (like the SD) the same. E(X+c)=E(X)+c SD(X+c)=SD(X) Var(X+c)=SD(X+c)2=SD(X)2=Var(X) For example: E(X+2)=E(X)+2 SD(X+2)=SD(X) Var(X+2)=Var(X) Rescaling will make both the measures of center (like mean) and the measures of spread (like the SD) to be rescaled. E(cX)=cE(X) SD(cX)=|c|SD(X) Var(cX)=SD(cX)2=|c|2 SD(X)2 =c2Var(X) c is an arbitrary constant (can be positive, negative or 0) For example: E(-2X)=-2E(X) SD(-2X)=2SD(X) Var(-2X)=4Var(X) More about E(X), Var(X) and SD(X) Shifting and Rescaling on a Random Variable If we consider the combination of the two transformations: shifting and rescaling, we can get the following results: E(aX+b)=E(aX)+b=aE(X)+b SD(aX+b)=SD(aX)=|a|SD(X) Var(aX+b)=Var(aX)=a2Var(X) For example: Consider the mean ,SD , and Variance of -2X+3 E(-2X+3)=-2E(X)+3 SD(-2X+3)=2SD(x) Var(-2X+3)=(-2)2Var(X)=4Var(X) More about E(X), Var(X) and SD(X) Let’s try an example together It’s determined that couples dining at the Quiet Nook can expect Lucky Lovers discounts averaging $5.83 with a standard deviation of $8.62. Suppose that for several weeks the restaurant has also been distributing coupons worth $5 off any one meal which can be used together with Lucky Lovers discounts. Question: If every couple dining there on the Valentine’s Day brings a coupon, what are the mean and standard deviation of the total discount each couple will receive? More about E(X), Var(X) and SD(X) Let’s try an example together Solution: Let X be the Lucky Lover discount each couple receive Then E(X)=$5.83, SD(X)=$8.62 Additionally with the $5 off any one meal coupon, the total discount=Lucky Lover discount+one meal diacount=X+5 Then E(X+5)=E(X)+5=5.83+5=$10.83 SD(X+5)=SD(X)=$8.62 More about E(X), Var(X) and SD(X) Let’s try an example together On the Valentine’s Day at the Quite Nook, couples may get a Lucky Lovers discount averaging $5.83 with a standard deviation of $8.62. When two couples dine together on a single check, the restaurant doubles the discount offer. Question: What are the mean and standard deviation of discounts for such couples? More about E(X), Var(X) and SD(X) Let’s try an example together Solution: Let X be the random variable of the Lucky Lover discount that one couple could receive. Then E(X)=$5.83 SD(X)=$8.62 When two couples dine on the same table, the total discount is 2X Then E(2X)=2E(X)=2*5.83=$11.66 and SD(2X)=2SD(X)=2*8.62=$17.24 More about E(X), Var(X) and SD(X) More than one variable For any two random variables X and Y E(X±Y)=E(X) ±E(Y) i.e. The expected value of the sum is the sum of the expected values For two independent random variables X and Y Var(X±Y)=Var(X) +Var(Y) Then More about E(X), Var(X) and SD(X) An application of the properties Let’s look at the following example: Jack and Sue want to compete their money investigation skills. Each of them will buy two stocks. Jack simply buy the two stocks from the same traded company F5 Networks. Based on the price history, the stock price for F5 Networks has the mean $22.09 and SD $ 7.82 However, Sue buy one stock from F5 Networks and the other one from the First Horizon company, an independent company from F5 Networks, whose history price mean and SD are the same as those of F5 Networks. Question: Whose strategy is better? More about E(X), Var(X) and SD(X) An application of the properties Solution: Let X be the random variable of the F5 stock price Let Y be the random variable of the F.H. stock price E(X)=E(Y)=$22.09 SD(X)=SD(Y)=$7.82 X and Y are independent random variables For Jack, total stock price=2X Since the SD for E(2X)=2E(X)=2*$22.09=$44.18 Sue’s stocks is SD(2X)=2SD(X)=2*$7.82=$15.64 smaller, we say For Sue, total stock price=X+Y Sue’s strategy is better because her E(X+Y)=E(X)+E(Y)=$22.09+$22.09=$44.18 stock price is more stable. More about E(X) and Var(X) • Let’s try the following example together. A grocery supplier believes that in a dozen eggs, the mean number of broken ones is 0.6 with a standard deviation of 0.5 eggs. You buy 3 dozen eggs without checking them. Assume the cartons are independent with each other. Questions: 1) What is the expected number of broken eggs? 2) What is the standard deviation of the number of the broken eggs? More about E(X) and Var(X) • Let’s try the following example together. In a litter of six kittens, two are female. You pick two kittens at random Questions: 1) Create a probability model for the number of male kittens you get. 2) What is the expected number of males you get? 3) What is the standard deviation ?