1 / 21

COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman

COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick. Chapter 6 : Probability. Key Terms and Formulas: Don’t Forget Notecards. Probability (p. 165) Random Sample (p. 167) Independent Random Sample (p. 167)

deiondre
Download Presentation

COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. COURSE: JUST 3900 TIPS FOR APLIA Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick Chapter 6: Probability

  2. 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:

  3. 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, • 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?

  4. Random Sampling • Question 1 Answer: • p = 17/25 = 0.68 • p = 16/25 = 0.64 • p = 4/25 = 0.16

  5. Random Sampling With and Without Replacement • Question 2:A jar contains 25 red marbles and 15 blue marbles. • 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?

  6. Random Sampling With and Without Replacement • Question 2 Answer: • 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.

  7. 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. • 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) = __

  8. Probability and Frequency Distributions • Question 3 Answer: • p(X = 4) = 4/8 = 0.500 • p(X < 4) = 3/8 = 0.375 • p(X > 4) = 1/8 = 0.125

  9. 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 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 ___.

  10. Properties of the Normal Curve

  11. Properties of the Normal Curve • Question 4 Answer: • 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. 2.15 + 13.59 + 34.13 + 34.13 + 13.59 + 2.15 = 99.74% 75 80 85 95 65 70 90

  12. 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%. 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 75 80 85 70 90 95

  13. Properties of the Normal Curve • Question 4 Answer: • 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 80 85 75 90 95

  14. 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) • z < 0.28 • z > 0.84 • z > -1.25 • z < -1.85

  15. The Unit Normal Table • Question 5 Answer: • p = 0.6103 • p = 0.2005 • p = 0.8944 • p = 0.0322 z < 0.28 z > 0.84 z > -1.25 z < -1.85

  16. 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?

  17. Binomial Data • Question 6 Answer: • Find µ and σ. • Find z. • Use unit normal table. • 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.

  18. 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?

  19. Binomial Data • Question 7 Answer: • Find µ and σ. • Find z. • Use the unit normal table to find the proportion between z and the mean for each z-value. • p(X = 18) = p(z = ±0.17) = 0.0675 + 0.0675 = 0.1350 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.

  20. 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: • 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.

  21. Frequently Asked Questions FAQs • A few things to keep in mind about binomial distributions: • Binomial distributions work with discrete variables, but the normal distribution is continuous. However, binomial distributions approximatethe 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. A score of 1 spans From 0.5 to 1.5. A score of 10 spans from 9.5 to 10.5.

More Related