Report

COURSE: JUST 3900 INTRODUCTORY STATISTICS FOR CRIMINAL JUSTICE Chapter 6: Probability Peer Tutor Slides Instructor: Mr. Ethan W. Cooper, Lead Tutor © 2013 - - PLEASE DO NOT CITE, QUOTE, OR REPRODUCE WITHOUT THE WRITTEN PERMISSION OF THE AUTHOR. FOR PERMISSION OR QUESTIONS, PLEASE EMAIL MR. COOPER AT THE FOLLWING: coopere07@students.ecu.edu Key Terms and Formulas: Don’t Forget Notecards Probability (p. 165) Random Sample (p. 167) Independent Random Sample (p. 167) Binomial Distribution (p. 185) Binomial Formulas: Mean: = Standard Deviation: = z-Score: = − = − Random Sampling Question 1: A survey of students in a criminal justice class revealed that there are 17 males and 8 females. Of the 17 males, only 5 had no brothers or sisters, and 4 of the females were also the only child in the household. If a student is randomly selected from this class, a) b) c) What is the probability of obtaining a male? What is the probability of selecting a student who has at least one brother or sister? What is the probability of selecting a female who has no siblings? Random Sampling Question 1 Answer: a) b) c) p = 17/25 = 0.68 p = 16/25 = 0.64 p = 4/25 = 0.16 Random Sampling With and Without Replacement Question 2: A jar contains 25 red marbles and 15 blue marbles. a) b) c) If you randomly select 1 marble from the jar, what is the probability of obtaining a red marble? If you take a random sample of n = 3 marbles from the jar and the first two marbles are both blue, what is the probability that the third marble will be red? If you take a sample (without replacement) of n = 3 marbles from the jar and the first two marbles are both red, what is the probability that the third marble will be blue? Random Sampling With and Without Replacement Question 2 Answer: a) b) c) p = 25/40 = 0.625 p = 25/40 = 0.625 p = 15/38 = 0.395 Remember that random sampling requires sampling with replacement. Here, we did not replace the first two red marbles that were drawn. Probability and Frequency Distributions Question 3: Consider the following frequency distribution histogram for a population that consists of N = 8 scores. Suppose you take a random sample of one score from this set. a) b) c) The probability that this score is equal to 4 is p(X = 4) = ____ The probability that this score is less than 4 is p(X < 4) = ____ The probability that this score is greater than 4 is p(X > 4) = __ Probability and Frequency Distributions Question 3 Answer: a) b) c) p(X = 4) = 4/8 = 0.500 p(X < 4) = 3/8 = 0.375 p(X > 4) = 1/8 = 0.125 Properties of the Normal Curve Question 4: The scores for students on Dr. Anderson’s research methods test had a mean of µ = 80 and a standard deviation of σ = 5. Use the figure on the next slide to answer the following questions. a) b) c) A score of 65 is ___ standard deviations below the mean, while a score of 95 is ___ standard deviations above the mean. This means that the percentage of students with scores between 65 and 95 is ___. A score of 90 is ___ standard deviations above the mean. As a result, the percentage of students with scores below 90 is ___. You can infer that 84.13% of students have scores above ___. Properties of the Normal Curve Properties of the Normal Curve Question 4 Answer: a) A score of 65 is _3_ standard deviations below the mean, while a score of 95 is _3_ standard deviations above the mean. This means that the percentage of students with scores between 65 and 95 is _99.74%. Add the percentages between -3σ and +3σ. 65 2.15 + 13.59 + 34.13 + 34.13 + 13.59 + 2.15 = 99.74% 70 75 80 85 90 95 Properties of the Normal Curve Question 4 Answer: A score of 90 is _2_ standard deviations above the mean. As a result, the percentage of students with scores below 90 is 97.72%. b) 13.59 + 34.13 + 34.13 + 13.59 + 2.15 + 0.13 = 97.72% Score of 90. or 100 – 2.15 – 0.13 = 97.72% 65 70 75 80 85 90 95 Properties of the Normal Curve Question 4 Answer: c) You can infer that 84.13% of students have scores above _75_. Start from 100 and subtract until you reach 84.13%. 84.13 % of students scored above a 75. 100 – 0.13 – 2.15 – 13.59 – 34.13 - 34.13 = 84.13% 65 70 75 80 85 90 95 The Unit Normal Table Question 5: Use the unit normal table (p. 699) to find the proportion of a normal distribution that corresponds to each of the following sections: (Hint: Make a sketch) a) b) c) d) z < 0.28 z > 0.84 z > -1.25 z < -1.85 The Unit Normal Table Question 5 Answer: a) b) c) d) p = 0.6103 p = 0.2005 p = 0.8944 p = 0.0322 z < 0.28 z > 0.84 z > -1.25 z < -1.85 Binomial Data Question 6: In the game Rock-Paper-Scissors, the probability that both players will select the same response and tie is p = 1/3, and the probability that they will pick different responses is q = 2/3. If two people play 72 rounds of the game and choose there responses randomly, what is the probability that they will choose the same response (tie) more than 28 times? Binomial Data Question 6 Answer: 1) 2) Find µ and σ. 1) = = 2) = = ∗ (72) = 24 1 3 72 ∗ 2 ∗ ( 3) = 16 = 4 Find z. 1) 3) 1 3 = − = − = 28.5−24 4 = 4.5 4 = 1.13 Use unit normal table. 1) p(X > 28.5) = p(z > 1.13) = 0.1292. Don’t forget real limits. We’re looking for the probability Of MORE than 28. Hence, we Use the upper real limit of 28.5. Binomial Data Question 7: If you toss a balanced coin 36 times, you would expect, on the average, to get 18 heads and 18 tails. What is the probability of obtaining exactly 18 heads in 36 tosses? Binomial Data Question 7 Answer: 1) 2) 3) Find µ and σ. 1) = = 0.5 ∗ (36) = 18 2) = = 36 ∗ 0.5 ∗ (0.5) = 9=3 Find z. 1) = − = − = 17.5−18 3 = −0.50 3 2) = − = − = 18.5−18 3 = 0.50 3 = −0.17 = 0.17 Don’t forget to use real limits. X = 18 spans the interval from 17.5 to 18.5. Therefore, we have to find the z-score for both the upper and lower real limits. Use the unit normal table to find the proportion between z and the mean for each z-value. 1) p(X = 18) = p(z = ±0.17) = 0.0675 + 0.0675 = 0.1350 Frequently Asked Questions FAQs How does one know if a question is asking for random sampling with replacement or random sampling without replacement? Unless the question specifically states that the sample was taken without replacement, always assume that the sample took place with replacement. Remember the requirements for random samples: 1) 2) Every individual in the population must have an equal chance of being selected. The probability of being selected must stay constant from one selection to the next if more than one individual is being selected. Frequently Asked Questions FAQs A few things to keep in mind about binomial distributions: A score of 1 spans From 0.5 to 1.5. A score of 10 spans from 9.5 to 10.5. Binomial distributions work with discrete variables, but the normal distribution is continuous. However, binomial distributions approximate the normal distribution when pn and qn are both greater than or equal to 10. But keep in mind that each X value actually corresponds to bar in the histogram. Therefore, a score of 10 is bounded by the real limits of 9.5 and 10.5.