Developed By Ethan Cooper Lead Tutor John Lohman Michael Mattocks Aubrey Urwick Chapter 6 Probability Key Terms and Formulas Dont Forget Notecards Probability p 165 Random Sample p 167 ID: 649266
Download Presentation The PPT/PDF document "COURSE: JUST 3900 TIPS FOR APLIA" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
COURSE: JUST 3900TIPS FOR APLIADeveloped By: Ethan Cooper (Lead Tutor) John LohmanMichael MattocksAubrey Urwick
Chapter 6
:
ProbabilitySlide2
Key Terms and Formulas: Don’t Forget NotecardsProbability (p. 165)Random Sample (p. 167)Independent Random Sample (p. 167)Binomial Distribution (p. 185)Binomial Formulas:Mean: Standard Deviation: z-Score:
Slide3
Random SamplingQuestion 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?Slide4
Random SamplingQuestion 1 Answer:p = 17/25 = 0.68p = 16/25 = 0.64p = 4/25 = 0.16Slide5
Random Sampling With and Without ReplacementQuestion 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?Slide6
Random Sampling With and Without ReplacementQuestion 2 Answer:p = 25/40 = 0.625p = 25/40 = 0.625p = 15/38 = 0.395Remember that random sampling requires sampling with replacement.Here, we did not replace the first two red marbles that were drawn.Slide7
Probability and Frequency DistributionsQuestion 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) = __Slide8
Probability and Frequency DistributionsQuestion 3 Answer:p(X = 4) = 4/8 = 0.500p(X < 4) = 3/8 = 0.375p(X > 4) = 1/8 = 0.125Slide9
Properties of the Normal CurveQuestion 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 ___.Slide10
Properties of the Normal CurveSlide11
Properties of the Normal CurveQuestion 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%.80859095
75
70
65
Add the percentages
b
etween -3
σ
and +3
σ
.
2.15 + 13.59 + 34.13 +
34.13 + 13.59 + 2.15 =
99.74%Slide12
Properties of the Normal CurveQuestion 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%.80856570
75
90
95
Score of 90.
13.59 + 34.13 + 34.13 +
13.59 + 2.15 + 0.13 =
97.72%
or
100 – 2.15 – 0.13 = 97.72%Slide13
Properties of the Normal CurveQuestion 4 Answer:You can infer that 84.13% of students have scores above _75_.65707580
85
90
95
84.13 % of students
s
cored above a 75.
Start from 100 and subtract
until you reach 84.13%.
100 – 0.13 – 2.15 – 13.59 –
34.13 - 34.13 = 84.13%Slide14
The Unit Normal TableQuestion 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 < 0.28z > 0.84z > -1.25z < -1.85Slide15
The Unit Normal TableQuestion 5 Answer:p = 0.6103p = 0.2005p = 0.8944p = 0.0322z < 0.28
z > 0.84
z > -1.25
z < -1.85Slide16
Binomial DataQuestion 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?Slide17
Binomial DataQuestion 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.Slide18
Binomial DataQuestion 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?Slide19
Binomial DataDon’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.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
Slide20
Frequently Asked Questions FAQsHow 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.Slide21
Frequently Asked Questions FAQsA few things to keep in mind about binomial distributions: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.A score of 10 spans from 9.5 to 10.5.
A score of 1 spans
From 0.5 to 1.5.