Math 1 Quiz: Relative Frequencies
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Relative FrequenciesQuestion 1 of 9

A company tracks employee satisfaction by department and years of experience. Among employees with more than 5 years of experience, 40% are satisfied. Among employees with 5 or fewer years of experience, 70% are satisfied. If 60% of all employees have more than 5 years of experience, what is the relative frequency of satisfied employees among all employees?

0.52 or 52% of all employees are satisfied
0.55 or 55% of all employees are satisfied
0.66 or 66% of experienced employees are satisfied
0.58 or 58% of all employees are satisfied
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Math 1 Quiz

Math 1 Quiz: Relative Frequencies

Practice Relative Frequencies in Math 1 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 Relative Frequencies, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

A company tracks employee satisfaction by department and years of experience. Among employees with more than 5 years of experience, 40% are satisfied. Among employees with 5 or fewer years of experience, 70% are satisfied. If 60% of all employees have more than 5 years of experience, what is the relative frequency of satisfied employees among all employees?

  1. 0.52 or 52% of all employees are satisfied (correct answer)
  2. 0.55 or 55% of all employees are satisfied
  3. 0.66 or 66% of experienced employees are satisfied
  4. 0.58 or 58% of all employees are satisfied
Explanation: This requires computing overall relative frequency from conditional frequencies. Let total employees = 1. Experienced (>5 years): 0.6, with 0.4 satisfied = 0.6 × 0.4 = 0.24 satisfied. New (≤5 years): 0.4, with 0.7 satisfied = 0.4 × 0.7 = 0.28 satisfied. Total satisfied: 0.24 + 0.28 = 0.52. Choice B incorrectly averages 0.4 and 0.7. Choice C gives a conditional rather than overall frequency. Choice D uses incorrect weights.

Question 2

A health clinic recorded patient visits by insurance type and appointment outcome. The data shows that 85% of insured patients kept their appointments, while only 40% of uninsured patients kept their appointments. If patients who kept appointments represent 75% of all scheduled appointments, what percentage of all scheduled patients were insured?

  1. Approximately 85.0% of all scheduled patients
  2. Approximately 65.0% of all scheduled patients
  3. Approximately 77.8% of all scheduled patients (correct answer)
  4. Approximately 70.0% of all scheduled patients
Explanation: This is a conditional probability problem that requires you to work backwards from given percentages to find what portion of patients were insured. When you see problems involving overlapping groups with different success rates, set up equations using the total outcomes. Let's define variables: let pp = proportion of patients who are insured, so (1−p)(1-p) = proportion uninsured. Since 75% of all patients kept appointments, we can write: 0.75=p×0.85+(1−p)×0.400.75 = p \times 0.85 + (1-p) \times 0.40 This equation says: total kept appointments = (insured patients × their success rate) + (uninsured patients × their success rate). Solving: 0.75=0.85p+0.40−0.40p0.75 = 0.85p + 0.40 - 0.40p 0.75=0.45p+0.400.75 = 0.45p + 0.40 0.35=0.45p0.35 = 0.45p p=0.350.45=79≈0.778p = \frac{0.35}{0.45} = \frac{7}{9} ≈ 0.778 So approximately 77.8% of patients were insured. Looking at the wrong answers: A) 85.0% incorrectly assumes the insured percentage equals the insured success rate. B) 65.0% might result from calculation errors or misunderstanding which percentages to use. D) 70.0% could come from averaging the given percentages (85% and 40% gives 62.5%, or including 75% gives roughly 70%), but this ignores the weighted relationship. Study tip: In weighted average problems, the final result will be closer to whichever group is larger. Since 77.8% is much closer to 85% than 40%, this confirms that most patients were insured, making our answer reasonable.

Question 3

A marketing survey classified consumers by age group and product preference. Among consumers who prefer Product A, 30% are young adults (18-35). Among young adults, 45% prefer Product A. If young adults represent 40% of all survey participants, what is the relative frequency of consumers who prefer Product A?

  1. 30.0% of all survey participants prefer Product A
  2. 60.0% of all survey participants prefer Product A (correct answer)
  3. 45.0% of all survey participants prefer Product A
  4. 37.5% of all survey participants prefer Product A
Explanation: Let A = prefers Product A, Y = young adults. Given: P(Y|A) = 0.30, P(A|Y) = 0.45, P(Y) = 0.40. From P(A|Y) × P(Y) = P(A∩Y) = P(Y|A) × P(A): 0.45 × 0.40 = 0.30 × P(A). Solving: 0.18 = 0.30 × P(A), so P(A) = 0.18/0.30 = 0.60 or 60%. Choices A, C, D represent direct misuse of the given conditional probabilities without proper calculation.

Question 4

A survey of 240 college students examined their exercise habits and stress levels. The results showed that 60 students exercise regularly and have low stress, 90 students exercise regularly and have high stress, 30 students don't exercise regularly and have low stress, and 60 students don't exercise regularly and have high stress. If a student is selected at random from those who have high stress, what is the probability that this student exercises regularly?

  1. 38\frac{3}{8}
  2. 35\frac{3}{5} (correct answer)
  3. 924\frac{9}{24}
  4. 12\frac{1}{2}
Explanation: This requires conditional probability using relative frequencies. First, find the total number of high-stress students: 90 + 60 = 150. Among these 150 high-stress students, 90 exercise regularly. So the probability is 90/150 = 3/5. Choice A incorrectly uses 90/240. Choice C uses the wrong fraction 9/24 from misreading the data. Choice D assumes equal distribution without calculating.

Question 5

A university studied the relationship between student housing type and academic performance. The analysis revealed that 70% of students in dormitories achieved a GPA above 3.0, while 55% of off-campus students achieved a GPA above 3.0. If students with GPA above 3.0 represent 65% of the entire student body, what percentage of students live in dormitories?

  1. Approximately 55.0% of all students live in dormitories
  2. Approximately 70.0% of all students live in dormitories
  3. Approximately 66.7% of all students live in dormitories (correct answer)
  4. Approximately 62.5% of all students live in dormitories
Explanation: When you encounter problems about overlapping groups with different success rates, you're dealing with weighted averages. The key insight is that the overall percentage (65%) must fall between the two subgroup percentages (70% and 55%), and its exact position depends on how the population is divided. Let's set up the problem systematically. If we let xx represent the fraction of students living in dormitories, then (1−x)(1-x) represents the fraction living off-campus. The weighted average equation becomes: 0.70x+0.55(1−x)=0.650.70x + 0.55(1-x) = 0.65 Expanding: 0.70x+0.55−0.55x=0.650.70x + 0.55 - 0.55x = 0.65 Simplifying: 0.15x=0.100.15x = 0.10 Therefore: x=0.100.15=23≈0.667x = \frac{0.10}{0.15} = \frac{2}{3} ≈ 0.667 or 66.7% Answer A (55.0%) incorrectly assumes the dormitory percentage equals the off-campus success rate, confusing the two different measurements. Answer B (70.0%) mistakenly uses the dormitory success rate as the population percentage, again mixing up success rates with population distribution. Answer D (62.5%) appears to come from averaging the two success rates (70% and 55%), but this ignores the constraint that the overall success rate must be 65%. The correct answer is C (66.7%). Strategy tip: In weighted average problems, always set up an equation where each subgroup's contribution equals its size times its rate. The answer will always make mathematical sense—here, having more students in the higher-performing group (dormitories) logically pushes the overall average closer to 70%.

Question 6

A restaurant analyzed customer orders by meal type and payment method. Among lunch customers, 65% paid with credit card. Among dinner customers, 80% paid with credit card. If lunch customers represent 70% of all credit card transactions, what proportion of all customers ordered lunch?

  1. Approximately 72.5% of all restaurant customers
  2. Approximately 65.0% of all restaurant customers
  3. Approximately 70.0% of all restaurant customers
  4. Approximately 74.2% of all restaurant customers (correct answer)
Explanation: When you encounter problems mixing percentages from different groups, you need to work backwards from the given relationships to find the underlying proportions. This is a classic conditional probability setup. Let's define our variables: Let LL = proportion of all customers who order lunch, so (1−L)(1-L) = proportion who order dinner. We know that 65% of lunch customers use credit cards, 80% of dinner customers use credit cards, and lunch customers make up 70% of all credit card transactions. The key insight is setting up an equation based on credit card usage. The total credit card transactions from lunch customers equals 0.65L0.65L, and from dinner customers equals 0.80(1−L)0.80(1-L). Since lunch customers represent 70% of all credit card transactions: 0.65L0.65L+0.80(1−L)=0.70\frac{0.65L}{0.65L + 0.80(1-L)} = 0.70 Solving this equation: 0.65L=0.70[0.65L+0.80(1−L)]0.65L = 0.70[0.65L + 0.80(1-L)] 0.65L=0.455L+0.56(1−L)0.65L = 0.455L + 0.56(1-L) 0.65L=0.455L+0.56−0.56L0.65L = 0.455L + 0.56 - 0.56L 0.65L=0.56−0.105L0.65L = 0.56 - 0.105L 0.755L=0.560.755L = 0.56 L=0.742L = 0.742 Therefore, approximately 74.2% of all customers ordered lunch, making D correct. Option A (72.5%) likely comes from incorrectly averaging the percentages. Option B (65.0%) mistakenly uses just the lunch credit card rate. Option C (70.0%) incorrectly assumes the proportion of lunch customers equals their share of credit card transactions. Remember: when dealing with conditional percentages across groups, always set up equations that account for the relative sizes of each group, not just the rates within groups.

Question 7

A medical study tracked 400 patients by age group (Under 50/50 and Over) and treatment outcome (Improved/No Change). The study found that 65% of all patients improved, 45% of patients are under 50, and the relative frequency of improvement among patients under 50 is 0.80. Among patients who did not improve, what percentage are 50 and over?

  1. 64.3%
  2. 71.4% (correct answer)
  3. 57.1%
  4. 42.9%
Explanation: Under 50: 180 patients, with 144 improved and 36 no change. 50 and over: 220 patients, with 116 improved and 104 no change. Total no change = 140. Among those with no change, 104 are 50 and over: 104/140 = 0.714 = 71.4%. Choice A uses wrong denominator (all patients). Choice C gives the percentage of under 50 among no change patients. Choice D represents the complement of choice C, not the correct answer.

Question 8

A transportation survey categorized 600 commuters by transportation method (Car/Public Transit) and commute distance (Under 10 miles/10+ miles). The data showed that 55% use cars, 40% commute under 10 miles, and among those who commute under 10 miles, 25% use public transit. What is the relative frequency of short-distance commuters (under 10 miles) among those who use public transit?

  1. 29\frac{2}{9} (correct answer)
  2. 13\frac{1}{3}
  3. 518\frac{5}{18}
  4. 718\frac{7}{18}
Explanation: Under 10 miles: 240 commuters, with 60 using transit and 180 using cars. 10+ miles: 360 commuters, with 210 using transit and 150 using cars. Total transit users: 270. Among transit users, 60 commute under 10 miles, so relative frequency = 60/270 = 2/9. Choice B assumes equal distribution. Choice C uses incorrect transit user counts. Choice D represents a calculation error in the conditional probability setup.

Question 9

A company surveyed employees about job satisfaction (Satisfied/Not Satisfied) and department (Sales/Engineering). The data shows that 70% of all employees are satisfied, 60% work in Sales, and among Sales employees, 80% are satisfied. What is the relative frequency of Engineering employees among those who are not satisfied?

  1. 12\frac{1}{2}
  2. 23\frac{2}{3} (correct answer)
  3. 13\frac{1}{3}
  4. 35\frac{3}{5}
Explanation: Let total = 100. Sales = 60, Engineering = 40. Satisfied Sales = 48, Not satisfied Sales = 12. Total satisfied = 70, so satisfied Engineering = 22, not satisfied Engineering = 18. Total not satisfied = 30. Among the 30 not satisfied, 18 are Engineering, so relative frequency = 18/30 = 2/3. Choice A assumes equal distribution. Choice C gives the proportion of Sales among not satisfied. Choice D uses incorrect calculation of the engineering proportions.