Random Variables Probability Continued Chapter 6 Random Variables

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Random Variables Probability Continued Chapter 6

Random Variables Probability Continued Chapter 6

Random Variables Suppose that each of three randomly selected customers purchasing a hot tub

Random Variables Suppose that each of three randomly selected customers purchasing a hot tub at a certain store chooses either an electric (E) or a gas (G) model. Assume that these customers make their choices independently of one another and that 40% of all customers select an electric model. The number among the three customers who purchase an electric hot tub is a random variable. What is the probability distribution?

Random Variable Example X = number of people who purchase electric hot tub X

Random Variable Example X = number of people who purchase electric hot tub X 0 1 2 3 P(X). 216. 432. 288. 064 GGG (. 6)(. 6) EGG GEG GGE (. 4)(. 6)(. 4) EEG GEE EGE (. 4)(. 6)(. 4) EEE (. 4)(. 4)

Random Variables A numerical variable whose value depends on the outcome of a chance

Random Variables A numerical variable whose value depends on the outcome of a chance experiment is called a random variable. discrete versus continuous

Discrete vs. Continuous The number of desks in a classroom. The fuel efficiency (mpg)

Discrete vs. Continuous The number of desks in a classroom. The fuel efficiency (mpg) of an automobile. The distance that a person throws a baseball. The number of questions asked during a statistics final exam.

Discrete versus Continuous Probability Distributions Which is which? Properties: For every possible x value,

Discrete versus Continuous Probability Distributions Which is which? Properties: For every possible x value, 0 < x < 1. l Sum of all possible probabilities add to 1. l Properties: Often represented by a graph or function. l Area of domain is 1. l

Means and Variances The mean value of a random variable X (written mx )

Means and Variances The mean value of a random variable X (written mx ) describes where the probability distribution of X is centered. We often find the mean is not a possible value of X, so it can also be referred to as the “expected value. ” The standard deviation of a random variable X (written sx )describes variability in the probability distribution.

Mean of a Random Variable Example Below is a distribution for number of visits

Mean of a Random Variable Example Below is a distribution for number of visits to a dentist in one year. X = # of visits to the dentist. Determine the expected value, variance and standard deviation.

Formulas Mean of a Random Variable Variance of a Random Variable

Formulas Mean of a Random Variable Variance of a Random Variable

Mean of a Random Variable Example E(X) = 0(. 1) + 1(. 3) +

Mean of a Random Variable Example E(X) = 0(. 1) + 1(. 3) + 2(. 4) + 3(. 15) + 4(. 05) = 1. 75 visits to the dentist

Variance and Standard Deviation of a Random Variable Example Var(X) = (0 – 1.

Variance and Standard Deviation of a Random Variable Example Var(X) = (0 – 1. 75)2(. 1) + (1 – 1. 75)2(. 3) + (2 – 1. 75)2(. 4) + (3 – 1. 75)2(. 15) + =. 9875 (4 – 1. 75)2(. 05)

Developing Transformation Rules Consider the following distribution for the random variable X:

Developing Transformation Rules Consider the following distribution for the random variable X:

X+1 What is the probability distribution for X+1?

X+1 What is the probability distribution for X+1?

2 X What is the probability distribution for 2 X?

2 X What is the probability distribution for 2 X?

Consider Suppose that E(X) = 2. 5, Var(X) = 0. 2 What is E(X+5)

Consider Suppose that E(X) = 2. 5, Var(X) = 0. 2 What is E(X+5) = ? , Var(X+5) = ? E(X+5) = 2. 5 + 5 = 7. 5 Var(X+5) = 0. 2 (no change) What is E(X – 2. 2) = ? , Var(X – 2. 2) = ? E(X – 2. 2) = 2. 5 – 2. 2 = 0. 3 Var(X – 2. 2) = 0. 2 (no change)

Consider Suppose that E(X) = 2. 5, Var(X) = 0. 2 What is E(3

Consider Suppose that E(X) = 2. 5, Var(X) = 0. 2 What is E(3 X) = ? , Var(3 X) = ? E(3 X) = 3*2. 5 = 7. 5 Var(3 X) = 32 * 0. 2 = 1. 8 What is E(2 X – 1) = ? , Var(2 X – 1) = ? E(2 X – 1) = 2(2. 5) – 1 = 4 Var(2 X – 1) = 22 * 0. 2 = 0. 8

Rule 1: If X is a random variable and a and b are fixed

Rule 1: If X is a random variable and a and b are fixed numbers, then ma + b. X = a + bm. X Rule 1: If X is a random variable and a and b are fixed numbers, then s 2 a + b. X =b 2 s 2 X

X+X What is the probability distribution for X+X? . 2 . 8 1 .

X+X What is the probability distribution for X+X? . 2 . 8 1 . 2 1 . 8 2 . 2 1 2. 8 2

X+X

X+X

X–X What is the probability distribution for X–X ? . 2 . 8 1

X–X What is the probability distribution for X–X ? . 2 . 8 1 . 2 1 . 8 2 . 2 1 2. 8 2

X–X

X–X

Consider Suppose that E(X) = 2. 5, Var(X) =. 16, E(Y) = 1. 2,

Consider Suppose that E(X) = 2. 5, Var(X) =. 16, E(Y) = 1. 2, Var(Y) =. 36 What is E(X+Y) = ? , Var(X+Y) = ? E(X+Y) = 2. 5 + 1. 2 = 3. 7 Var(X+Y) =. 16 +. 36 =. 52 E(X – Y) = ? , Var(X – Y) = ? E(X – Y) = 2. 5 – 1. 2 = 1. 3 Var(X – Y) =. 16 +. 36 =. 52

Consider E(X) = 2. 5, Var(X) =. 16, E(Y) = 1. 2, Var(Y) =.

Consider E(X) = 2. 5, Var(X) =. 16, E(Y) = 1. 2, Var(Y) =. 36 What is s(X) = ? , s(Y) = ? , s(X+Y) = ? s(X) =. 4, s(Y) =. 6 Since Var(X+Y) =. 52, s(X+Y) =. 7211 l Cannot add standard deviation directly!! What is E(2 X – 4 Y) = ? , s(2 X-4 Y) = ? E(2 X– 4 Y) = 2(2. 5) – 4(1. 2) = 0. 2 Var(2 X– 4 Y) = 22(. 16) + 42(. 36) = 6. 4 So s(2 X– 4 Y) = 2. 53

Rule 2: If X and Y are random variables, then m. X + Y

Rule 2: If X and Y are random variables, then m. X + Y = m. X + m. Y m. X – Y = m. X – m. Y Rule 2: If X and Y are independent random variables, then s 2 X + Y = s 2 X + s 2 Y s 2 X - Y = s 2 X + s 2 Y Note: cannot combine standard deviation directly.