/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 Five hundred first-year students... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Five hundred first-year students at a state university were classified according to both high school GPA and whether they were on academic probation at the end of their first semester. The data are $$ \begin{array}{lcccc} && {\text { High School GPA }} \\ & 2.5 \text { to } & 3.0 \text { to } & 3.5 \text { and } & \\ \text { Probation } & <3.0 & <3.5 & \text { Above } & \text { Total } \\ \hline \text { Yes } & 50 & 55 & 30 & 135 \\ \text { No } & 45 & 135 & 185 & 365 \\ \text { Total } & 95 & 190 & 215 & 500 \\ \hline \end{array} $$ a. Construct a table of the estimated probabilities for each GPA-probation combination. b. Use the table constructed in Part (a) to approximate the probability that a randomly selected first-year student at this university will be on academic probation at the end of the first semester. c. What is the estimated probability that a randomly selected first-year student at this university had a high school GPA of \(3.5\) or above? d. Are the two outcomes selected student has a bigh school GPA of \(3.5\) or above and selected student is on academic probation at the end of the first semester independent outcomes? How can you tell? e. Estimate the proportion of first-year students with high school GPAs between \(2.5\) and \(3.0\) who are on academic probation at the end of the first semester. f. Estimate the proportion of those first-year students with high school GPAs \(3.5\) and above who are on academic probation at the end of the first semester.

Short Answer

Expert verified
a. See Step 1 for the table of estimated probabilities. b. The probability for a student to be on probation is 0.27. c. The probability for a student to have a high school GPA of 3.5 or above is 0.43. d. The two outcomes selected student has a high school GPA of 3.5 or above and selected student is on academic probation at the end of the first semester are not independent outcomes. (See Step 4 for calculation) e. The proportion of first-year students with high school GPAs between 2.5 and 3.0 and who are on academic probation at the end of the first semester is 0.53. f. The proportion of those first-year students with high school GPAs 3.5 and above and who are on academic probation at the end of the first semester is 0.14.

Step by step solution

01

Construct the table of estimated probabilities

Each cell within the table represents the joint probability of the two outcomes \(P(A \cap B) = \frac{n(A \cap B)}{n(S)}\) where \(n(A \cap B)\) represents the intersection count and \(n(S)\) represents the total count. For instance, the probability of a student with a GPA of less than 3.0 being on probation is \(\frac{50}{500} = 0.1\). All the other probabilities can be calculated in a similar way.
02

Probablity for a student to be on probation

The probability of a student being on probation \(P(Probation) = \frac{n(Probation)}{n(S)}\). From the table, we know that the total number of students on probation is 135. So, the probability will be \(\frac{135}{500} = 0.27\).
03

Probablity for a student to have a high school GPA of 3.5 or above

Probability \(P(GPA \geq 3.5) = \frac{n(GPA \geq 3.5)}{n(S)}\). From the table, we can see that the total number of students with a GPA of 3.5 and above is 215, hence \(\frac{215}{500} = 0.43\).
04

Check for independence of two events

Two events A & B are said to be independent if \(P(A \cap B) = P(A)P(B)\). We compute \(P(GPA \geq 3.5 \cap Probation) = P(GPA \geq 3.5)P(Probation)\). If these values are equal, the events are independent; otherwise, they are dependent.
05

Estimate the proportion of first-year students with high school GPAs between 2.5 and 3.0

For students with GPAs between 2.5 and 3.0, we can calculate the proportion of students on probation as \(P(Probation | GPA = 2.5-3.0) = \frac{n(Probation \cap GPA = 2.5-3.0)}{n(GPA = 2.5-3.0)} = \frac{50}{95} = 0.53\).
06

Estimate the proportion of those first-year students with high school GPAs 3.5 and above

For students with GPAs 3.5 and above, we can calculate the proportion of students on probation as \(P(Probation | GPA \geq 3.5) = \frac{n(Probation \cap GPA \geq 3.5)}{n(GPA \geq 3.5)} = \frac{30}{215} = 0.14 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Joint Probability
Joint Probability refers to the probability of two events occurring simultaneously. It helps in understanding how likely two scenarios are to happen at the same time. In our exercise, we look at students' high school GPA and whether they are on academic probation simultaneously. Here's how we assess joint probability:
  • We determine it by dividing the frequency of both events happening by the total number of observations.
  • This is the probability of a specific GPA bracket and being on probation occurring together.
For example, there are 50 students with a GPA below 3.0 on probation. The joint probability is \[P\left(GPA < 3.0 \cap \text{Probation}\right) = \frac{50}{500} = 0.1 \]Hence, there's a 10% chance a random student will both have a GPA less than 3.0 and be on probation.
Conditional Probability
Conditional Probability is the likelihood of an event occurring given that another event has occurred. It provides insights into how one condition affects the occurrence of another condition, which is essential in checking dependencies between events. It's especially useful in analyzing patterns within datasets.For instance, if we want to know the probability of a student being on probation given they have a GPA between 2.5 and 3.0, we use the formula:\[P(\text{Probation} | GPA = 2.5-3.0) = \frac{P(GPA = 2.5-3.0 \cap \text{Probation})}{P(GPA = 2.5-3.0)}\]Calculation:- Number on probation (GPA 2.5-3.0) = 50- Total in GPA range 2.5-3.0 = 95\[P(\text{Probation} | GPA = 2.5-3.0) = \frac{50}{95} \approx 0.53\]Therefore, there's approximately a 53% chance a student with a GPA between 2.5 and 3.0 will be on probation.
Independence of Events
The concept of independence between two events in probability suggests that the occurrence of one event does not affect the occurrence of the other. For two events A and B to be independent, the following must hold true:\[P(A \cap B) = P(A)P(B)\]In our example, we are examining whether having a GPA of 3.5 and above is independent of being on academic probation.Check for Independence:
  • Calculate the individual probability of GPA ≥ 3.5:
  • \[P(GPA \geq 3.5) = \frac{215}{500} = 0.43\]
  • Calculate the individual probability of being on probation:
  • \[P(\text{Probation}) = \frac{135}{500} = 0.27\]
  • Calculate the joint probability of GPA ≥ 3.5 and on probation:
  • \[P(GPA \geq 3.5 \cap \text{Probation}) = \frac{30}{500} = 0.06\]Since 0.43 * 0.27 ≠ 0.06, the events are not independent. Therefore, GPA ≥ 3.5 does influence the likelihood of probation.
High School GPA Analysis
Analyzing students' high school GPA can provide valuable insights into their academic journeys and likelihood of facing probation. By breaking down GPAs into categories, institutions can offer targeted support to students who may be at risk. Analysis Steps:
  • We categorize students based on their high school GPA ranges: below 3.0, between 3.0-3.5, and above 3.5.
  • We then assess how many in each category are on academic probation.
Findings: For students with a GPA:
  • < 3.0, roughly 52.6% (50 out of 95) are on probation.
  • Between 3.0 and 3.5, roughly 28.9% (55 out of 190) are on probation.
  • 3.5 and above, only about 14% (30 out of 215) are on probation.
This pattern suggests a trend where students with lower high school GPAs are more susceptible to academic probation, highlighting the importance of early support interventions.

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Most popular questions from this chapter

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. a. Suppose that the hosts randomly select two answers from those submitted with replacement. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

Suppose that a box contains 25 light bulbs, of which 20 are good and the other 5 are defective. Consider randomly selecting three bulbs without replacement. Let \(E\) denote the event that the first bulb selected is good, \(F\) be the event that the second bulb is good, and \(G\) represent the event that the third bulb selected is good. a. What is \(P(E)\) ? b. What is \(P(F \mid E)\) ? c. What is \(P(G \mid E \cap F)\) ? d. What is the probability that all three selected bulbs are good?

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\), and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F\) ? b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F)\) ? A Venn diagram might help.)

The paper "Good for Women, Good for Men, Bad for People: Simpson's Paradox and the Importance of Sex-Spedfic Analysis in Observational Studies" (Journal of Women's Health and Gender-Based Medicine [2001]: \(867-872\) ) described the results of a medical study in which one treatment was shown to be better for men and better for women than a competing treatment. However, if the data for men and women are combined, it appears as though the competing treatment is better. To see how this can happen, consider the accompanying data tables constructed from information in the paper. Subjects in the study were given either Treatment \(\mathrm{A}\) or Treatment \(\mathrm{B}\), and survival was noted. Let \(S\) be the event that a patient selected at random survives, \(A\) be the event that a patient selected at random received Treatment \(\mathrm{A}\), and \(B\) be the event that a patient selected at random received Treatment \(\mathrm{B}\). a. The following table summarizes data for men and women combined: $$ \begin{array}{l|ccc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 215 & 85 & \mathbf{3 0 0} \\ \text { Treatment B } & 241 & 59 & \mathbf{3 0 0} \\ \text { Total } & \mathbf{4 5 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? b. Now consider the summary data for the men who participated in the study: $$ \begin{array}{l|rrr} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 120 & 80 & \mathbf{2 0 0} \\ \text { Treatment B } & 20 & 20 & 40 \\ \text { Total } & \mathbf{1 4 0} & \mathbf{1 0 0} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? c. Now consider the summary data for the women who participated in the study: $$ \begin{array}{l|rrc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 95 & 5 & \mathbf{1 0 0} \\ \text { Treatment B } & 221 & 39 & \mathbf{2 6 0} \\ \text { Total } & \mathbf{3 1 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? d. You should have noticed from Parts (b) and (c) that for both men and women, Treatment \(A\) appears to be better. But in Part (a), when the data for men and women are combined, it looks like Treatment \(\mathrm{B}\) is better. This is an example of what is called Simpson's paradox. Write a brief explanation of why this apparent inconsistency occurs for this data set. (Hint: Do men and women respond similarly to the two treatments?)

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let \(n o t E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(E_{1}\) )? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

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