Textbook Notes (368,831)
QMS 102 (49)
Chapter 5-8

# QMS 102 Chapters 5-8 Review Sheet.pdf

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School
Department
Quantitative Methods
Course
QMS 102
Professor
Clare Chua
Semester
Winter

Description
Chapter 5 – Basic Probability Probability is determining the probability that a particular event will occur. Probability of occurrence = X / T where X = the number of ways in which a particular event occurs and T = the total number of possible outcomes. Joint probability refers to the probability of an occurrence involving two or more events. The marginal probability of an event consists of a set of joint probabilities. Two events are mutually exclusive if both the events cannot occur simultaneously. Conditional probability refers to the probability of event A, given information about the occurrence of another event B. Example 1 (joint probability): Find the probability that a randomly selected household that purchased a big-screen television also purchased a plasma-screen television and a DVR. Purchased DVR Purchased Plasma Yes No Total Screen Plasma Screen 38 42 80 Not Plasma Screen 70 150 220 Total 108 192 300 P (Purchased Screen and DVR) = number of people that purchased a plasma screen and a DVR / total number of big-screen television purchasers  from the table, we can see that 38 people purchased a plasma screen and a DVR. P (Purchased Screen and DVR) = 38 / 300 = 0.127 or 12.7%. Example 2 (marginal probability): Referring to example 1, find the probability of the households that purchased a plasma screen or a DVR. P (Plasma Screen or DVR) = P (Plasma Screen) + P (Purchased DVR) – P (Plasma Screen and DVR)  we have to remember to subtract the probability of those who purchased the plasma screen and the DVR. P (Purchased Screen or DVR = (80 / 100) + (108 / 300) – (38/300) = 150 / 300 = 0.5 or 50%. Example 3 (conditional probability): Referring to example 1, find the probability of a household that purchased a DVR, given that the household purchased a plasma-screen television. P (Purchased DVR | Purchased Plasma Screen) = number of people that purchased a plasma screen and DVR / number of people that purchased a plasma screen P (Purchased DVR | Purchased Plasma Screen = 38 / 80 = 0.475 or 47.5% Another way of solving this question would be through using the formula P (B | A) = P (A and B) / P (A) where A = number of people that purchased a plasma and B = number of people that purchased a DVR. P (B | A) = (38 / 300) / (80 / 300) = 0.475 or 47.5%. Example 4 (multiplication rule): Referring to example 1, suppose three households are randomly selected from the 80 households. Find the probability that all three households purchased a DVR. P (A and B) = (38 / 80) x (37 / 79) x (36 / 78) = 0.1027 or 10.27%. Notice how both the numerator and denominator both decrease by one because after we select the first household, there are only 37 people left who purchased a DVR and 79 people left in the total sample. Chapter 6 – Discrete Probability Distributions Random variable: a variable whose value is determined by the outcome of a random experiment. Discrete random variable: a random variable whose possible values are discrete or individual values. Discrete probability distribution: a table that shows the probability associated with each possible value of a discrete random variable. Mean (µ) = sample (n) x probability (p)  µ = np Variance (σ ) = [n x p x (1-p)] Standard deviation (σ) = square root of [n x p x (1-p)] or the square root of the variance Example 1: Find the probability distribution and the expected value of tossing a fair coin three times. Let us first list the total possible outcomes or our sample space and remember the total of all the probabilities has to equal 1 or 100%. S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} and let X = the number of tails. X P(X) 0 1 / 8 = 0.125 1 3 / 8 = 0.375 2 3 / 8 = 0.375 3 1 / 8 = 0.125 Now that we have the probability distribution, we have to find the expected value, which we can do one of two ways. The first method is through multiplying X by P(X) and adding all the sums together. µ = E(X) = (0)(0.125) + (1)(0.375) + (2)(0.375) + (3)(0.125) = 1.5 and the second method involves putting the data from X into list 1 on your calculator and inputting the data from P(X) into list 2 and then calculating the mean through the CALC and 1-variable functions. Example 2: A company has determined that the probability of a unit being successful in Brampton is 3/4 and the annual profit in this case is \$150,000, but if unsuccessful, there will be losses of \$80,000. In Mississauga, the probability of attaining a profit of \$240,000 is 1/2, but the potential losses amount to \$48,000. Where should the company locate to maximize expected profit and which of the two locations is riskier? Brampton Mississauga Profit Prob. Profit Prob. \$150,000 3 / 4 = 0.75 \$240,000 2 / 4 = 0.50 -\$80,000 1 / 4 = 0.25 -\$48,000 2 / 4 = 0.50 Step 1) Enter Profit into List 1 Step 2) Enter Prob. into List 2 Step 3) CALC  1-VAR mean = \$92,500 (expected profit) and standard deviation = \$99,592.92 for Brampton mean = \$96,000 (expected profit) and standard deviation = \$144,000 for Mississauga Now we have to determine the coefficient of variation to determine which location is riskier. CV Brampton (σ / µ) x 100 = (\$99,592.92 / \$92,500) x 100 = 107.67% CV Mississaugaσ / µ) x 100 = (\$144,000 / \$96,200) x 100 = 150% Therefore, the company should locate itself in Mississauga because it has a higher expected profit of \$96,000, but it is also riskier as the coefficient of variation is equal to 150%. Example 3 (binomial distribution): If 100 people are sampled and it is determined that 40% of them prefer to pay their telephone bills online, what is the probability that at least 30 customers paid their telephone bills online? Since each trial is independent, we can use the binomial distribution to solve for the probability. If we use the Bpd function, it would only tell us an exact number of customers so we have to use the Bcd function to calculate the probability over a certain number of customers to calculate our answer. DIST  BINM  Bcd (x = 29, n = 100, p = 0.40) = P (X ≤ 29), but we want the P (X ≥ 30) so we have to subtract 1 - P (X ≤ 29) = P (X ≥ 30) and the calculator always tells us ≤ (less than or equal to) a certain #. P (X ≤ 29) = 0.01477531  P (X ≥ 30) = 1 – 0.01477351 = 0.98522469 Type 10 (an exact More than/at Less than 10 At most or no number) least/less than 10 more than 10 Bpd X Bcd X – subtract from 1 X (we would use X 9 in the calculator) *If we are asked to calculate more than 10 or a certain number depending on the question, we have to subtract 1 – P (X ≤ 10) because the calculator only tells us the probability of the left side or less than and equal to 10 or whatever the number so the probability of P (X ≥ 10) would be found by 1 – P (X ≤ 9). Example 4 (poisson distribution): A computer in Mark’s hardware store breaks down an average of 2 times per month. What is the probability that the computer will break down 15 times this year? We have no sample of computers provided to us in this question nor are we giv
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