
Concept explainers
To find: The value of joint distribution, marginal distribution and the conditional distribution for the first and second scenario.

Answer to Problem 21E
Solution: Joint Distribution for first scenario is:
Time Period | |||
Design | More than a minute | Less than a minute | Total |
Design1 | 24.00% | 26.00% | 50.00% |
Design2 | 10.00% | 40.00% | 50.00% |
Total | 34.00% | 66.00% | 100.00% |
The conditional distribution is:
Design | More than a minute | Less than a minute |
Design1 | 70.58% | 39.4% |
Design2 | 29.42% | 60.6% |
Total | 100.00% | 100.00% |
And,
Time Period | |||
Design | More than a minute | Less than a minute | Total |
Design1 | 48.00% | 52.00% | 100.00% |
Design2 | 20.00% | 80.00% | 100.00% |
Now, joint distribution for second scenario is:
Response | |||
Student Type | Yes | No | Total |
1st year | 14.60% | 47.40% | 62.00% |
4th year | 20.10% | 17.90% | 38.00% |
Total | 34.70% | 65.30% | 100.00% |
The conditional distribution is,
Student Type | More than a minute | Less than a minute |
1st year | 42.00% | 72.60% |
4th year | 58.00% | 27.40% |
Total | 100.00% | 100.00% |
And,
Response | |||
Student Type | Yes | No | Total |
1st year | 23.50% | 76.45% | 100.00% |
4th year | 53.00% | 47.00% | 100.00% |
Explanation of Solution
Given: In the study, for the first scenario, there are 12 students who look at the 1st design for more than a minute and 5 students who look at the 2nd design for the same time period. But, there are 13 students, who look at 1st design for less than a minute and 20 students who look at 2nd design for less than a minute.
The 2 x 2 table for first scenario is as follows:
Time Period | |||
Design | More than a minute | Less than a minute | Total |
Design1 | 12 | 13 | 25 |
Design2 | 5 | 20 | 25 |
Total | 17 | 33 | 50 |
For the second scenario, there are two types of student, one in the 1st year and other in the 4th year and their responses to the newly proposed core curriculum. Among the 361, 1st year students, 85 respond yes and 276 respond no. While, among the 221, 4th year students, 117 responded yes and 104 responded no.
The 2 x 2 table for the second scenario is as follows:
Response | |||
Student Type | Yes | No | Total |
1st year | 85 | 276 | 361 |
2nd year | 117 | 104 | 221 |
Total | 202 | 380 | 582 |
Calculations:
In the study, for the first scenario, the Joint distribution is computed by dividing the cell element by the total observation. The obtained joint distribution is shown below:
Time Period | |||
Design | More than a minute | Less than a minute | Total |
Design1 | |||
Design2 | |||
Total |
Now, the marginal distribution is computed by dividing the row or column totals by the overall total. Marginal distributions provide information about the individual variables but not about the relationship between two variables.
Thus, the marginal distribution of designs is shown below:
Design | Marginal Distribution |
Design1 | |
Design2 |
And the marginal distribution of time duration is shown below:
Time Period | ||
More than a minute | Less than a minute | |
Marginal probability |
Conditional distribution is obtained by dividing the row or column elements by the sum of the observations in the corresponding row or column. The conditional distribution of Time Period by Design is shown below:
Time Period | |||
Design | More than a minute | Less than a minute | Total |
Design1 | |||
Design2 |
The conditional distribution of Design by Time Period is shown below:
Design | More than a minute | Less than a minute |
Design1 | ||
Design2 | ||
Total |
For the second scenario, the calculations are as follows:
In the study, Joint distribution is computed by dividing the cell element by the total observation. The obtained joint distribution is shown below:
Response | |||
Student Type | Yes | No | Total |
1st year | |||
2nd year | |||
Total |
Now, the marginal distribution is computed by dividing the row or column totals by the overall total. Marginal distributions provide information about the individual variables but not about the relationship between two variables. Thus, the marginal distribution for the types of students is shown below:
Student Type | Marginal Distribution |
1st year | |
2nd year |
Whereas, the marginal distribution of responses is shown below:
Response | ||
Yes | No | |
Marginal Probability |
Conditional distribution is obtained by dividing the row or column elements by the sum of the observations in the corresponding row or column.
The conditional distribution of Responses by Student Type is shown below:
Response | |||
Student Type | Yes | No | Total |
1st year | |||
4th year |
The conditional distribution of Student Type by Responses is shown below:
Student Type | More than a minute | Less than a minute |
1st year | ||
4th year | ||
Total |
Interpretation: The above 2x2 tables for the first scenario show different percentage values for joint, marginal and conditional distributions. The percentage of looking at the 2nd design for more than a minute is 10% and the percentage of looking at the 2nd design for less than a minute is 40%. The value of the marginal distribution for more than a minute is 34% and for less than a minute is 66%. The conditional distributional for the percentage of Time Period by Design for looking at the 1st design for more than a minute is 48%, the percentage of looking at the 2nd design for the same is 20%.
Also, for the second scenario, the percentage of 4th year students responding yes and no for the new curriculum are 20.1% and 17.9% respectively. The marginal distribution for the responding yes and no are 34.7% and 65.3% respectively. The conditional probability of responding no by the 1st year student is 76.45%.
Want to see more full solutions like this?
Chapter 9 Solutions
Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
- An electronics company manufactures batches of n circuit boards. Before a batch is approved for shipment, m boards are randomly selected from the batch and tested. The batch is rejected if more than d boards in the sample are found to be faulty. a) A batch actually contains six faulty circuit boards. Find the probability that the batch is rejected when n = 20, m = 5, and d = 1. b) A batch actually contains nine faulty circuit boards. Find the probability that the batch is rejected when n = 30, m = 10, and d = 1.arrow_forwardTwenty-eight applicants interested in working for the Food Stamp program took an examination designed to measure their aptitude for social work. A stem-and-leaf plot of the 28 scores appears below, where the first column is the count per branch, the second column is the stem value, and the remaining digits are the leaves. a) List all the values. Count 1 Stems Leaves 4 6 1 4 6 567 9 3688 026799 9 8 145667788 7 9 1234788 b) Calculate the first quartile (Q1) and the third Quartile (Q3). c) Calculate the interquartile range. d) Construct a boxplot for this data.arrow_forwardPam, Rob and Sam get a cake that is one-third chocolate, one-third vanilla, and one-third strawberry as shown below. They wish to fairly divide the cake using the lone chooser method. Pam likes strawberry twice as much as chocolate or vanilla. Rob only likes chocolate. Sam, the chooser, likes vanilla and strawberry twice as much as chocolate. In the first division, Pam cuts the strawberry piece off and lets Rob choose his favorite piece. Based on that, Rob chooses the chocolate and vanilla parts. Note: All cuts made to the cake shown below are vertical.Which is a second division that Rob would make of his share of the cake?arrow_forward
- Three players (one divider and two choosers) are going to divide a cake fairly using the lone divider method. The divider cuts the cake into three slices (s1, s2, and s3). If the choosers' declarations are Chooser 1: {s1 , s2} and Chooser 2: {s2 , s3}. Using the lone-divider method, how many different fair divisions of this cake are possible?arrow_forwardTheorem 2.6 (The Minkowski inequality) Let p≥1. Suppose that X and Y are random variables, such that E|X|P <∞ and E|Y P <00. Then X+YpX+Yparrow_forwardTheorem 1.2 (1) Suppose that P(|X|≤b) = 1 for some b > 0, that EX = 0, and set Var X = 0². Then, for 0 0, P(X > x) ≤e-x+1²² P(|X|>x) ≤2e-1x+1²² (ii) Let X1, X2...., Xn be independent random variables with mean 0, suppose that P(X ≤b) = 1 for all k, and set oσ = Var X. Then, for x > 0. and 0x) ≤2 exp Σ k=1 (iii) If, in addition, X1, X2, X, are identically distributed, then P(S|x) ≤2 expl-tx+nt²o).arrow_forward
- Theorem 5.1 (Jensen's inequality) state without proof the Jensen's Ineg. Let X be a random variable, g a convex function, and suppose that X and g(X) are integrable. Then g(EX) < Eg(X).arrow_forwardCan social media mistakes hurt your chances of finding a job? According to a survey of 1,000 hiring managers across many different industries, 76% claim that they use social media sites to research prospective candidates for any job. Calculate the probabilities of the following events. (Round your answers to three decimal places.) answer parts a-c. a) Out of 30 job listings, at least 19 will conduct social media screening. b) Out of 30 job listings, fewer than 17 will conduct social media screening. c) Out of 30 job listings, exactly between 19 and 22 (including 19 and 22) will conduct social media screening. show all steps for probabilities please. answer parts a-c.arrow_forwardQuestion: we know that for rt. (x+ys s ا. 13. rs. and my so using this, show that it vye and EIXI, EIYO This : E (IX + Y) ≤2" (EIX (" + Ely!")arrow_forward
- Theorem 2.4 (The Hölder inequality) Let p+q=1. If E|X|P < ∞ and E|Y| < ∞, then . |EXY ≤ E|XY|||X|| ||||qarrow_forwardTheorem 7.6 (Etemadi's inequality) Let X1, X2, X, be independent random variables. Then, for all x > 0, P(max |S|>3x) ≤3 max P(S| > x). Isk≤narrow_forwardTheorem 7.2 Suppose that E X = 0 for all k, that Var X = 0} x) ≤ 2P(S>x 1≤k≤n S√2), -S√2). P(max Sk>x) ≤ 2P(|S|>x- 1arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin HarcourtGlencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill

