1. CONTINUOUS RANDOM VARIABLES Continuous random variables take all values in an interval Too many values! Impossible to list them all! How to describe a continuous random variable? Use density curve. P(a<X<b) abX P(a ≤X < b)=area under the density curve between a and b. Total area under density curve =1. Density curve of random variable X Values of X

2. Continuous density example -uniform distribution X-time (min) taken by a fire truck to arrive at a fire after 911 call. Suppose X is equally well anywhere between (2, 15) min. X r.v. with uniform distribution on (2, 15). Density curve for X has the same height over (2, 15): rectangle. Area=1. Length (2, 15) is 13 height=1/13. 1/13 02 5815 P(5<X<8) Density of X Probability that it takes between 5 and 8 min for the fire truck to arrive: P( 5<X < 8) = Area between 5 and 8=3 x (1/13)=3/13.

3. MEAN AND VARIANCE OF A RANDOM VARIABLE Measures of center and spread of a r.v. X – discrete r.v. X has probability distribution: Center of the distribution: mean, Expected value of X, EX or E(X): Values of the r.v X x1x1 x2x2 …xnxn probabilitiesp1p1 p2p2 …pnpn NOTES: The mean represents the “long-run-average” value of X. If we average many values of X, we expect to get a number close to EX. EX is a weighted average of the values of X, weights are the probabilities of the values.

4. VARIANCE and STANDARD DEVIATION OF A RANDOM VARIABLE Measures of spread around the mean μ of a random variable. X discrete rv Variance of X Take squared deviations from the mean and add them up with the same weights as for the mean: VarX=Var(X)= σ 2 = (x 1 – μ) 2 x p 1 + (x 2 – μ) 2 x p 2 + … + (x n – μ) 2 x p n = Standard deviation of X: NOTES: Both variance and standard deviation of any random variable are nonnegative. If X is continuous use calculus to define and compute mean and variance. NO WORRIES IN STAT152.

5. EXAMPLE The number of defective chips produced in a month by a company ABCHIP has the following probability distribution: Number of defective chips0101520 probabilities0.10.20.50.2 a. What is the average number of defective chips this company produces in a month? b. What is the standard deviation of the number of defective chips this company produces in a month? Solution. a. X=# of defective chips produced by ABCHIP in a month. EX= μ = 0 x 0.1 + 10 x 0.2 + 15 x 0.5 + 20 x 0.2 = 13.5 Note that the mean is not one of the possible values of X! b. VarX= σ 2 = (0-13.5) 2 x 0.1 + (10 – 13.5) 2 x 0.2 + (15 – 13.5) 2 x 0.5+(20-13.5) 2 x 0.2 = 30.25. Standard deviation of X,

6. EXAMPLE Your winnings on a lottery are a random variable with the following probability distribution: Winnings in $1002001,0000 Probabilities0.40.3?0.2 a.Find the probability that you win $1,000. b. What is the probability that you will win at least $150? c.What is your expected winning? Solution. X= winnings in $. a.Since sum of all probs=1, then P(X=1,000)=1-(0.4+0.3+0.2)=0.1. b. P(X ≥ $150)=P(X=$200 or X=$1,000)=0.3 +0.1=0.4. c. EX = 0 x 0.2 + 100 x 0.4 + 200 x 0.3 +1,000 x 0.1= 200. You should expect to win about $200 if you play for a long time.

7. EXAMPLE What if we consider that you must pay $10 to play the lottery. What are your expected winning then? The probability distribution of the winnings changes to accommodate the payment. Winnings in $90190990-10 Probabilities0.40.30.10.2 The negative $10 means you lose $10 with probability 0.2. EX= -10 x 0.2 + 90 x 0.4 + 190 x 0.3 +990 x 0.1 = 190. You should expect to win about $190 if you play for a long time.