Chapter 14: random variables p394 A random variable (r. v.) is a variable whose value is a numerical outcome of a random phenomenon. Consider the experiment of tossing a coin. Define a random variable as follows X = 1 if a H comes up = 0 if a T comes up. - This is an example of a Bernoulli r.v. Probability function of X x P(X = x) 0 1 q p p +q = 1 1
Probability distributions Each value of a random variable is an event, so each value has probability. List of values and probabilities called probability model. Tossing 3 coins: # heads 0 1 2 3 Prob. 1 3 3 1 8 8 8 8 2
Combining values of random variable: 3 coins: # heads 0 1 2 3 Prob. 1 3 3 1 8 8 8 8 How likely are we to get two or more heads? add up probs: 3/8+1/8=4/8=1/2 How likely to get at least one head? P(no heads)=1/8, so P(at least one)=1-1/8=7/8 or: P(1 or 2 or 3)=3/8+3/8+1/8=7/8 3
The mean of a random variable p394 Here's a random variable, called X: Value of X 2 3 4 5 Probability 0.1 0.2 0.4 0.3 Mean not (2+3+4+5)/4=3.5 because 4 and 5 more likely than 2 or 3. Have to account for more likely values when adding up: times by probability: 2(0.1)+3(0.2)+4(0.4)+5(0.3)=0.2+0.6 +1.6+1.5=3.9. (Weighted average, weights sum to 1.) Median is value of X where summed-up probabilities first pass 0.5: 3 too small (total 0.1+0.2=0.3), 4 is right (0.1+0.2+0.4=0.7), so median 4. Mean a little smaller than median: leftskewed. 4
The variance of a r. v. P397 - The variance of a r. v. is an average of the squared deviations ( X µ ) 2 X - Variance of a discrete r. v. is Var( X ) = ( x µ ) 2P( x) - The standard deviation of a r. v. is the positive square root of its variance. - Examples 5
Linear changes to a random variable What does it mean to add a to a random variable? Multiply it by b? Take all the values and change them, while leaving the probabilities alone. Here's Y, with mean 4 and SD 0.45: Value of Y 3 4 5 Probability 0.1 0.8 0.1 2Y looks like this. Check that mean now 8, SD 0.9. Value of Y 6 8 10 Probability 0.1 0.8 0.1 and Y+3 as below. Check that mean now 7, SD 0.45. Value of Y 6 7 8 Probability 0.1 0.8 0.1 6
Some useful results If you add a constant to a random variable, what happens to its mean? SD? Mean of (X+a) = mean of X plus a (i.e E(X+a)=E(X)+a SD of (X+a) = SD of X ( i.e. SD(X+a) = SD(X)) If you multiply a random variable by a constant, what happens to its mean? SD? Mean of bx = b times mean of X (i.e. E(aX)=a E(X)) SD of bx = b times SD of X. (i.e. SD(aX) =a SD(X)) 7
- For any two variables X and Y, - E(X+Y) = E(X) + E(Y) - E(X-Y) = E(X) - E(Y) If X and Y are independent, then SD( X + Y) = SD2( X ) + SD2( Y) SD( X Y) = SD2( X ) + SD2( Y) 8
Continuous random variables So far: our random variables discrete: set of possible values, like 1,2,3,..., probability for each. Recall normal distribution: any decimal value possible, can't talk about probability of any one value, just eg. less than 10, between 10 and 15, greater than 15. Normal random variable example of continuous. Finding mean and SD of continuous random variable involves calculus :-( but if we are given mean/sd, work as above. 9
Handling two normal distributions p413 Result If X N µ, ~ (, ( )) 1 X SD X 1 1 X ~ N( µ, SD( X )), and 2 X 2 2 are independent, then X and 1 X 2 X + ~ (, 2( ) 2( )) 1 X 2 N µ + X µ SD X + X 1 SD X 2 1 2 and X ~ (, 2( ) 2( )) 1 X 2 N µ X µ SD X + X 1 SD X 2 1 2. 10
Example The weight of the empty box has a normal distribution with mean 1kg and std. dev. 100g. The weight of its contents has a normal distribution with mean 12kg and std. dev. 1.34 kg, independently of the box. Find the probability that the total weight of the box and its contents will exceed 15kg. 11
Ex. Two friends T and H run a race. H is a faster runner and the time he takes to complete is normally distributed with mean 3 minutes with a std. dev. 30 sec. T s time to complete the race is normally distributed with mean 5 minutes and std. dev. 1 minute. Find the probability that T will win the race. Ans. P(T<H)= P(T-H<0)=P(Z<(0-(5-3))/sqrt(1.25))=P(Z<-1.79)= 0.0367 12
How do you find SD of sum and difference if random variables are not independent? In this course, you don't. See p. 404 of text for gory details. 13
Probability Models p 405 The Binomial Model Example: A biased coin (P(H) = p = 0.6) ) is tossed 5 times. Let X be the number of H s. Fine P(X = 2). This X is a binomial r. v. 14
The binomial setting 1. There are a fixed number n of observations. 2. The n observations are independent. 3. Each observation falls into one of just two categories (successes and failures) 4. The probability of a success (call it p) is the same for each observation. Probability function of the binomial dist. If X has a B(n, p), P( X = x) = nc x(1 ) n x x p p for n= 0,1,, n 15
Binomial table The link to Statistical Tables on course website includes table of binomial distribution probabilities. In here, find chance of exactly k successes in n trials with success prob p. Ex. The probability that a certain machine will produce a defective item is 1/5. If a random sample of 6 items is taken from the output of this machine, what is the probability that there will be 5 or more defectives in the sample? 16
Ex There are 20 multiple-choice questions on an exam, each having responses a, b, c, d and e. Each question is worth 5 points. And only one response per question is correct. Suppose that a student guesses the answer to question and her guesses from question to question are independent. It the student needs at least 40 points to pass the test. What is the probability that the student will pass the test? Ans. X~B(20, 0.2). P(X>=8) = 0.0322, adding the entries 8 through 20 in the appropriate of the binomial table What is the expected (mean) score for this student. (later) Ans. 20 x 0.2 = 4 and expected score =5 x 4= 20 17
Suppose n=8 and p=0.7. What is the probability of exactly 7 successes? 7 or more successes? Idea: count failures instead of successes. P(success)=0.7 means P(failure)=1-0.7=0.3 7 successes = 8-7=1 failure. so look up n=8, p=0.3, k=1 prob=0.1977 which is answer we want. 7 or successes = 7, 8 successes P(failure)=1-0.7=0.3 7, 8 successes = 1, 0 failures prob we want is 0.1977+0.0576=0.2553. 18
Mean and Variance of a binomial r. v. p If X has a Bin(n, p) mean np = and SD= np(1 p) Example 19
Binomial Distribution: Binomial trials=5, Probability of success=0.5 Probability Mass 0.05 0.10 0.15 0.20 0.25 0.30 0 1 2 3 4 5 Number of Successes 20
Binomial Distribution: Binomial trials=20, Probability of success=0.9 Probability Mass 0.00 0.05 0.10 0.15 0.20 0.25 13 14 15 16 17 18 19 20 Number of Successes Binomial Distribution: Binomial trials=30, Probability of success=0.2 Probability Mass 0.00 0.05 0.10 0.15 0 2 4 6 8 10 12 14 Number of Successes 21
Binomial Distribution: Binomial trials=500, Probability of success=0.4 Probability Mass 0.00 0.01 0.02 0.03 170 180 190 200 210 220 230 Number of Successes 22
How does the shape depend on p? p<0.5, skewed right; p>0.5, skewed left; p=0.5, symmetric What happens to the shape as n increases? shape becomes normal What does this suggest to do if n is too large for the tables? If n too large for tables, try normal approximation to binomial. Compute mean and SD of binomial, then pretend binomial actually normal: 23
Normal approximation for counts and proportions p415 Draw a SRS of size n from a large population having population p of success. Let X be the count of success in the sample and pˆ = X / n the sample proportion of successes. When n is large, the sampling distributions of these statistics are approximately normal: X is approx. N( np, np(1 p)) Works if n large and p not too far from 0.5 As a rule of thumb, we will use this approximation for values of n and p that satisfy np 10 and n(1 p) 10. can relax this a bit if p close to 0.5. 24
According to government data, 21% of American children under the age of six live in households with incomes less than the official poverty level. A study of learning in early childhood chooses an SRS of 300 children. (a) What is the mean number of children in the sample who come from poverty-level households? What is the standard deviation of this number? (b) Use the normal approximation to calculate the probability that at least 80 of the children in the sample live in poverty. Be sure to check that you can safely use the approximation. (a) µ = (300)(0.21) = 63, σ = ( 300)(0.21)(0.79) = 7.0548. (b) np = 63 and n(1 p) = 237 are both more than 10, so we may approximate using the normal distribution: P(X 80) = P(Z 2.41) = 0.0080, or with the continuity correction: P(X 79.5) = P(Z 2.34) = 0.0096. 25