Math 2 Quiz: Two Way Tables And Frequencies

12 questions · exam conditions

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Two Way Tables And FrequenciesQuestion 1 of 12

A company tracks employee satisfaction across three departments (Sales, Marketing, Operations) and two satisfaction levels (Satisfied, Not Satisfied). If the joint relative frequency for satisfied Sales employees is 0.15, the marginal relative frequency for Sales employees is 0.30, and there are 400 total employees, how many Sales employees are not satisfied?

45 employees who represent 0.1125 of the workforce

60 employees who represent 0.15 of the workforce

75 employees who represent 0.1875 of the workforce

90 employees who represent 0.225 of the workforce

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Math 2 Quiz

Math 2 Quiz: Two Way Tables And Frequencies

Practice Two Way Tables And Frequencies in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Two Way Tables And Frequencies, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

45 employees who represent 0.1125 of the workforce
60 employees who represent 0.15 of the workforce (correct answer)
75 employees who represent 0.1875 of the workforce
90 employees who represent 0.225 of the workforce

Explanation: Total Sales employees = 0.30 × 400 = 120. Satisfied Sales employees = 0.15 × 400 = 60. Not satisfied Sales employees = 120 - 60 = 60, which represents 60/400 = 0.15 of workforce. Choice A miscalculates the relative frequency. Choice C confuses joint and conditional frequencies. Choice D doubles the satisfied count incorrectly.

Question 2

A health clinic analyzed patient data for two treatments (Treatment A and Treatment B) and their outcomes (Improved, No Change, Worsened). The clinic wants to determine if there's an association between treatment type and patient outcomes.

If 40% of all patients received Treatment A, 25% of all patients improved, and the joint relative frequency for patients who received Treatment A and improved is 0.12, what is the conditional relative frequency that a patient improved given they received Treatment A?

0.12, indicating that 12% of Treatment A patients improved
0.30, indicating that 30% of Treatment A patients improved (correct answer)
0.48, indicating that 48% of Treatment A patients improved
0.52, indicating that 52% of Treatment A patients improved

Explanation: P(Improved | Treatment A) = P(Improved AND Treatment A) ÷ P(Treatment A) = 0.12 ÷ 0.40 = 0.30. Choice A confuses conditional with joint frequency. Choice C incorrectly multiplies probabilities (0.12 × 4). Choice D represents the complement incorrectly calculated.

Question 3

In constructing a two-way relative frequency table for student performance data, a researcher finds that the sum of joint relative frequencies in the first row is 0.45, the sum in the second row is 0.35, and the marginal relative frequency for the first column is 0.60. If the joint relative frequency for row 1, column 1 is $x$ , what is the joint relative frequency for row 2, column 1?

$0.60 - x$ , ensuring column totals are maintained correctly (correct answer)
$0.35 - x$ , based on the constraint from row 2 totals
$0.45 - x$ , derived from the first row total constraint
$1.00 - x$ , representing the complement of the given frequency

Explanation: Since the marginal frequency for column 1 is 0.60, and this equals the sum of all joint frequencies in that column, we have: (row 1, col 1) + (row 2, col 1) = 0.60. Therefore, (row 2, col 1) = 0.60 - x. Choice B uses row total instead of column total. Choice C confuses row and column constraints. Choice D uses total probability instead of column marginal.

Question 4

A researcher analyzing survey data notices that in a two-way table, the conditional relative frequency P(B|A) = 0.72 and the joint relative frequency P(A and B) = 0.288. Later, additional data changes P(A and B) to 0.324 while keeping the marginal relative frequency P(A) unchanged. What is the new conditional relative frequency P(B|A)?

0.90, representing the maximum possible increase given the constraints of the problem
0.64, representing a decrease caused by the change in joint frequency calculation
0.72, remaining unchanged since the marginal probability P(A) stayed constant throughout
0.81, representing an increase in conditional probability due to more joint occurrences (correct answer)

Explanation: When you encounter conditional probability problems involving changes to joint frequencies, remember that conditional probability depends on the relationship between joint and marginal probabilities through the formula

P(B|A) = \frac{P(A \text{ and } B)}{P(A)}

. First, let's find the original marginal probability P(A). Since

P(B|A) = \frac{P(A \text{ and } B)}{P(A)}

, we have

0.72 = \frac{0.288}{P(A)}

. Solving:

P(A) = \frac{0.288}{0.72} = 0.4

. When the joint probability changes to 0.324 while P(A) remains 0.4, the new conditional probability becomes:

P(B|A) = \frac{0.324}{0.4} = 0.81

. Answer A (0.90) incorrectly assumes this represents a maximum increase, but there's no such constraint given in the problem. Answer B (0.64) makes an error by somehow calculating a decrease when the joint frequency actually increased. Answer C (0.72) falls into the trap of thinking conditional probability stays constant when marginal probability is unchanged, but this ignores that conditional probability depends on both joint AND marginal frequencies. Answer D (0.81) correctly reflects the increase in conditional probability due to the higher joint frequency. Study tip: Always write out the conditional probability formula when solving these problems. The key insight is that even when P(A) stays constant, changes in P(A and B) will change P(B|A) proportionally. Practice identifying which probabilities are given and which need to be calculated.

Question 5

A quality control manager creates a two-way table for product defects with variables 'Shift' (Day, Night) and 'Defect Type' (Minor, Major, None). If P(Minor | Day) = 0.15, P(Major | Day) = 0.05, P(Minor | Night) = 0.25, and equal numbers of products are inspected on each shift, what is the joint relative frequency for products with no defects produced during the day shift?

0.50, representing all day shift products regardless of defect status
0.35, representing the proportion of defect-free day shift products overall
0.40, representing day shift products with no defects out of total production (correct answer)
0.80, representing day shift products with no defects as conditional frequency

Explanation: Two-way tables with conditional and joint probabilities require careful attention to what proportion you're calculating and relative to what total. When you see conditional probabilities like P(Minor | Day), these tell you the breakdown within each shift, but to find joint relative frequencies, you need to consider the entire population. Since equal numbers of products are inspected on each shift, each shift represents 50% of total production. For the day shift, you're given P(Minor | Day) = 0.15 and P(Major | Day) = 0.05. Since probabilities within a category must sum to 1, P(None | Day) = 1 - 0.15 - 0.05 = 0.80. This means 80% of day shift products have no defects. To find the joint relative frequency for day shift products with no defects, you multiply: P(Day and None) = P(Day) × P(None | Day) = 0.50 × 0.80 = 0.40. This represents 40% of all products inspected. Choice A (0.50) incorrectly gives you just the marginal probability of day shift production, ignoring defect status entirely. Choice B (0.35) appears to be a miscalculation, possibly confusing conditional and joint probabilities. Choice D (0.80) gives you the conditional probability P(None | Day) rather than the joint probability—this is the proportion of defect-free products within the day shift, not within total production. Remember: joint relative frequencies require multiplying the marginal probability by the conditional probability. Always check whether the question asks for a conditional probability (within a subgroup) or a joint probability (within the entire population).

Question 6

In a study comparing two teaching methods (Traditional, Interactive) and student outcomes (Excellent, Good, Fair), researchers find that P(Excellent | Interactive) = 0.45, P(Good | Interactive) = 0.35, and 60% of all students used the Interactive method. If the overall proportion of students achieving Excellent outcomes is 0.30, what is P(Excellent | Traditional)?

0.225, demonstrating Traditional method performs reasonably well for excellent outcomes
0.1875, showing Traditional method has moderate effectiveness for excellent outcomes
0.075, indicating Traditional method produces fewer excellent outcomes per student (correct answer)
0.30, showing Traditional method matches the overall excellent outcome rate exactly

Explanation: When you encounter conditional probability problems involving multiple groups, you need to use the law of total probability to connect the pieces. This question gives you conditional probabilities for one group and asks you to find the conditional probability for another group. Let's work through this systematically. You know that 60% of students used Interactive method, so 40% used Traditional. You can set up the law of total probability:

P(\text{Excellent}) = P(\text{Excellent | Interactive}) \times P(\text{Interactive}) + P(\text{Excellent | Traditional}) \times P(\text{Traditional})

Substituting the known values:

0.30 = 0.45 \times 0.60 + P(\text{Excellent | Traditional}) \times 0.40

Solving:

0.30 = 0.27 + P(\text{Excellent | Traditional}) \times 0.40

0.03 = P(\text{Excellent | Traditional}) \times 0.40

P(\text{Excellent | Traditional}) = 0.075

This means answer C is correct - the Traditional method produces fewer excellent outcomes per student. Looking at the wrong answers: A (0.225) would require the Traditional method to be quite effective, but this contradicts our calculation. B (0.1875) represents a moderate effectiveness level that's mathematically inconsistent with the given data. D (0.30) incorrectly assumes the Traditional method performs at the overall average rate, ignoring that the Interactive method's strong performance (0.45) must be balanced by weaker Traditional performance. Study tip: In conditional probability problems with multiple groups, always check that your answer makes intuitive sense - if one method significantly outperforms the average, the other must underperform to maintain the overall average.

Question 7

In a two-way table analyzing the relationship between study method (Online, In-Person) and test performance (Pass, Fail), the marginal relative frequency for students who passed is 0.75, and the conditional relative frequency for passing given online study is 0.60. If 40% of students studied online, what is the joint relative frequency for students who studied in-person and passed?

0.51, representing in-person students who passed the test (correct answer)
0.45, representing in-person students who passed the test
0.24, representing online students who actually passed the test
0.36, representing the total in-person student population

Explanation: Joint frequency (Online, Pass) = 0.40 × 0.60 = 0.24. Since total who passed = 0.75, then Joint frequency (In-Person, Pass) = 0.75 - 0.24 = 0.51. Choice B miscalculates using 0.75 × 0.60. Choice C gives the joint frequency for online students who passed. Choice D gives the marginal frequency for in-person students.

Question 8

A survey of 200 students asked about their preferred study method (online or in-person) and their grade level (freshman or sophomore). The results showed that 75% of freshmen prefer online study, while 40% of sophomores prefer online study. If there are equal numbers of freshmen and sophomores in the survey, what is the conditional relative frequency of being a freshman given that a student prefers in-person study?

0.25
0.30
0.29 (correct answer)
0.35

Explanation: First, set up the two-way table. With 100 freshmen and 100 sophomores: 75 freshmen prefer online (25 prefer in-person), and 40 sophomores prefer online (60 prefer in-person). Total preferring in-person = 25 + 60 = 85. The conditional relative frequency of being a freshman given preference for in-person study is 25/85 = 5/17 ≈ 0.29. Choice A uses marginal frequency (25/100). Choice B incorrectly uses 25/85 but rounds to 0.30. Choice D uses the wrong conditional (freshmen preferring in-person out of all freshmen).

Question 9

In a study of 300 patients, researchers found that among those who experienced side effects, 60% were taking medication A and 40% were taking medication B. Among those who did not experience side effects, 30% were taking medication A and 70% were taking medication B. If 80 patients total experienced side effects, what is the joint relative frequency of taking medication A and not experiencing side effects?

0.22 (correct answer)
0.24
0.26
0.28

Explanation: First, find the number of patients in each category. Side effects: 80 total, with 60% taking A = 48 patients. No side effects: 220 total, with 30% taking A = 66 patients. The joint frequency of medication A and no side effects is 66 patients. The joint relative frequency is 66/300 = 0.22. Choice B incorrectly uses 72/300. Choice C uses the marginal frequency of medication A. Choice D uses an incorrect calculation of the conditional frequencies.

Question 10

A researcher collected data on 350 participants regarding their exercise habits and stress levels. She found that the conditional relative frequency of high stress given regular exercise is 0.25, while the conditional relative frequency of high stress given irregular exercise is 0.60. If 40% of participants exercise regularly, what is the joint relative frequency of irregular exercise and low stress?

0.24 (correct answer)
0.30
0.36
0.42

Explanation: Regular exercisers = 40% = 140 people. Irregular exercisers = 60% = 210 people. Among irregular exercisers, P(high stress) = 0.60, so P(low stress) = 0.40. Number with irregular exercise and low stress = 210 × 0.40 = 84. Joint relative frequency = 84/350 = 0.24. Choice B uses the wrong percentage (0.30 instead of 0.40 for low stress). Choice C incorrectly uses regular exercisers. Choice D uses the total irregular exercise frequency.

Question 11

A survey of 480 voters examined the relationship between age group and voting preference. The data showed that among voters under 50, the ratio of those supporting Candidate A to those supporting Candidate B is 3:2. Among voters 50 and over, this ratio is 1:3. If there are equal numbers of voters in each age group, what is the joint relative frequency of being 50 or over and supporting Candidate B?

0.450
0.375 (correct answer)
0.300
0.325

Explanation: When you encounter problems involving joint relative frequency, you're looking at the proportion of the total population that falls into a specific intersection of two categories. Here, you need to find what fraction of all 480 voters are both "50 or over" AND "support Candidate B." Start by organizing the given information. With equal numbers in each age group, there are 240 voters under 50 and 240 voters 50 and over. For voters under 50, the 3:2 ratio means 3 parts support A and 2 parts support B, totaling 5 parts. So

\frac{2}{5} \times 240 = 96

under-50 voters support B. For voters 50 and over, the 1:3 ratio means 1 part supports A and 3 parts support B, totaling 4 parts. So

\frac{3}{4} \times 240 = 180

voters 50+ support B. The joint relative frequency of being 50+ AND supporting B is

\frac{180}{480} = 0.375

. Choice A (0.450) likely comes from incorrectly calculating the total proportion supporting B:

\frac{96 + 180}{480} = \frac{276}{480} = 0.575

, then making an arithmetic error. Choice C (0.300) might result from using

\frac{180}{600}

if you mistakenly thought there were 600 total voters. Choice D (0.325) could come from calculation errors when working with the ratios or confused fraction arithmetic. Remember: joint relative frequency always uses the total sample size as the denominator, and you must carefully track which specific intersection of categories you're calculating.

Question 12

A marketing firm surveyed 500 customers about their shopping preferences. They found that 45% prefer online shopping, and among online shoppers, 80% are under age 40. Among in-store shoppers, 35% are under age 40. What is the marginal relative frequency of customers under age 40?

0.54 (correct answer)
0.57
0.60
0.63

Explanation: Set up the two-way table: 45% (225) prefer online, 55% (275) prefer in-store. Among online shoppers: 80% under 40 = 225 × 0.80 = 180 customers. Among in-store shoppers: 35% under 40 = 275 × 0.35 = 96.25, which we round to 96 customers. Total under 40 = 180 + 96 = 276. Marginal relative frequency = 276/500 = 0.552 ≈ 0.55, closest to 0.54. Choice B uses an incorrect total. Choice C uses only the online percentage. Choice D miscalculates the in-store group.