6.9 – Discrete Random Variables: Mean and Standard Deviation IBHLY2 - Santowski (A) Mean of a Probability Distribution for a discrete random variable, X, which can assume the values x1, x2, x3, ....xn, the mean is also called the expected value of X on a very simple level, if there are n members in a given event, and each of the n members has a probability of occurring of p, then the expectation of occurrence of any specific event is n x p example if we roll a die, how many 6's are expected if we roll 120 times since the event of rolling 6's has a probability of occurring 1 in 6 times (i.e. p = 1/6) and we roll n = 120 times, it would be expected that we get the six 20 times (n x p = 120 x 1/6) (B) Class Work CLASSWORK to reinforce the idea of expected value: SL Math, Chap 29C, p715, Q1-9 HL Math, Chap 30C, p733, Q1-6 (C) Expected Value (Mean) by Formula the expected value formula comes from the general observation of n x p as before n now we write it as E ( X ) xi P( X xi ) i 1 and we use again as we are looking at the idea of a mean of a population xi represents specific outcomes P(X=xi) represents the probability of xi occurring (or pi) X represents the random variable we are concerned about the expected value represents the Along term average@ of the variable X. The effect of multiplying each value of x by its probability, p, gives it a weighting then summing all the weighted values gives us an overall expectation for X (C) Expected Value (Mean) by Formula ex 1 Find the mean of the probability distribution given in the table below: x 4 P(X = x) 0.002 6 8 10 0.040 0.299 0.659 What does the data mean? the specific outcome of 4 occurs with a probability of 0.002, the outcome 8 occurs with a probability of 0.299, the outcome 10 occurs with the highest probability of 0.659 = E(X) = xiP(X=xi) = [(4)(0.002) + (6)(0.040) + (8)(0.299) + (10)(0.659)] = 9.23 So the mean (expected value) is 9.23 so given the possible outcomes and their associated probabilities, you can expect a value of 9.23 (C) Expected Value (Mean) by Formula ex 2 Find the expected value of X, E(X), for the probability distribution given by the formula x 4 x 4 1 P( X x) x 3 2 3 , x 0,1,2,3,4 To interpret the formula we have 4 events occurring and the probability for “success” in any one event is 1/3 At times, the data is easier to work with, if we have a chart/table: X 0 P(X = x) 0.198 1 2 3 4 0.395 0.296 0.099 0.012 = E(X) = xiP(X=xi) = [(0)(0.197) + (1)(0.395) + (2)(0.296) + (3)(0.099) + (4)(0.012)] = 1.332 (C) Expected Value (Mean) by Formula Find the expected value of X, E(X), for the probability distribution given 4 3 x 3 x P( X x) , x 0,1,2,3 7 3 ANS = 1.713 (C) Expected Value (Mean) by Formula ex 4 A committee of 3 people is to be selected from 4 men and 2 women. Let X represent the number of women chosen. Find the expected value of X ANS = 1 (D) Homework SL Math Text, 29C, p716, Q10-14 HL Math Text, 30C, p734, Q7-9 Peter Smythe book, Mathematics SL&HL, Section 14.2, p391, Q1-8 (E) Variance and Standard Deviation recall that the way we calculated the population variance was to find the square of the deviation of each point from the population mean so in a similar manner, when we calculate the variance of a random variable, X, we wind up subtracting from our possible values of X (xi ) n Thus Var ( X ) E ( X ) ( xi ) 2 P( X xi ) 2 2 i 1 and SD( X ) Var ( X ) n (x ) i 1 i 2 P( X xi ) where = E(X) the standard deviation gives a measure of the way in which the values are spread about the mean (E) Variance and Standard Deviation other textbooks will express these formulas slightly differently, as they let pi represent P(X = xi) so the formulas are written as : n xi pi i 1 n ( xi ) pi 2 2 i 1 (E) Variance and Standard Deviation we can make one simplification to the variance formula 2 Var ( X ) 2 E ( x ) (x (x i ) pi 2 i pi ) 2 2 E( X ) 2 2 E ( X 2 ) [ E ( X )]2 (F) Examples ex 1 Find Var(X) and SD(X), given the probability distribution below: x 2 3 4 P(X = x) 0.3 0.5 0.2 We will set up a table to work through our calculation: x P(X=x) xP(X=x) x- (x- )2 (x- )2P(X=x) 2 0.3 0.6 -0.9 0.81 0.243 3 0.5 1.5 0.1 0.01 0.005 4 0.2 0.8 1.1 1.21 0.242 E( X ) xi P( X xi ) 29. = 0.490 (F) Examples therefore, in our example, Var(X) = E(x - u)2 = 0.490 and SD(X) = (0.490)0.5 = 0.7 now, using our alternative formula of Var(X) = E(X2) [E(X)]2, we first work out E(X2) as (x2)(P(X=x)) which is then 22(0.3) + 32(0.5) + 42(0.2) = 8.9 so we get Var(X) = 8.9 - 2.92 = 0.49, which is the same answer as when we used the complete table (G) Examples ex 2 Find the expected value (mean) and standard deviation for the random variable, X, which is the event of rolling a single die. so then 1 1 1 1 1 1 E ( X ) xi pi 1 2 3 4 5 6 3.5 6 6 6 6 6 6 2 2 2 1 2 1 2 1 2 1 2 1 2 1 E ( X ) xi pi 1 2 3 4 5 6 15 1 6 6 6 6 6 6 6 so then Var(X) = E(X2) - [E(X)]2 = 15.166666.... - (3.5)2 = 2.91666666... (G) Examples ex 3 Find the mean and standard deviation of the random variable, X, with the probability distribution given in the table: x 0 1 2 3 P(X=x) 0.264 0.494 0.220 0.022 (G) Examples ex 4 A debating team of 4 is to be chosen from 6 girls and 3 boys. Let X be the number of boys chosen. Find: (i) E(X) (ii) Var(X) (iii) SD(X) (H) Homework SL Math text, Chap 29D, p720, Q1-8 HL Math text, Chap 30D, p737, Q1-8