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6.9 – Discrete Random Variables: Mean and Standard Deviation

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 x 1 , x 2 , x 3 , .... x n , the mean is also called the expected value of X

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6.9 – Discrete Random Variables: Mean and Standard Deviation

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  1. 6.9 – Discrete Random Variables: Mean and Standard Deviation IBHLY2 - Santowski

  2. (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)

  3. (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

  4. (C) Expected Value (Mean) by Formula • the expected value formula comes from the general observation of n x p as before • now we write it as • 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

  5. (C) Expected Value (Mean) by Formula • ex 1  Find the mean of the probability distribution given in the table below: • 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

  6. (C) Expected Value (Mean) by Formula • ex 2  Find the expected value of X, E(X), for the probability distribution given by the formula • 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: •  = E(X) = xiP(X=xi) = [(0)(0.197) + (1)(0.395) + (2)(0.296) + (3)(0.099) + (4)(0.012)] = 1.332

  7. (C) Expected Value (Mean) by Formula • Find the expected value of X, E(X), for the probability distribution given • ANS = 1.713

  8. (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

  9. (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

  10. (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 - ) • Thus • and • where  = E(X) • the standard deviation gives a measure of the way in which the values are spread about the mean

  11. (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 :

  12. (E) Variance and Standard Deviation • we can make one simplification to the variance formula 

  13. (F) Examples • ex 1  Find Var(X) and SD(X), given the probability distribution below: • We will set up a table to work through our calculation:

  14. (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

  15. (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 • so then Var(X) = E(X2) - [E(X)]2 = 15.166666.... - (3.5)2 = 2.91666666...

  16. (G) Examples • ex 3  Find the mean and standard deviation of the random variable, X, with the probability distribution given in the table:

  17. (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)

  18. (H) Homework • SL Math text, Chap 29D, p720, Q1-8 • HL Math text, Chap 30D, p737, Q1-8

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