- A survey of 120 students asked about their preferred study location (Library or Home) and their class year (Freshman or Sophomore). The results showed that 45 freshmen preferred the library, 25 freshmen preferred home, 30 sophomores preferred the library, and 20 sophomores preferred home. If a student is randomly selected from those who prefer studying at home, what is the probability that the student is a freshman?
Answer: 95
First, identify the marginal frequency for students who prefer home: 25 freshmen + 20 sophomores = 45 total students prefer home. Among these 45 students who prefer home, 25 are freshmen. Therefore, P(Freshman|Home) = 4525=95. Choice B incorrectly uses total sophomores (50) instead of total who prefer home. Choice C gives the joint probability P(Freshman and Home). Choice D uses an incorrect denominator calculation. - A researcher collected data on 150 participants' exercise habits (Regular or Occasional) and diet type (Vegetarian or Non-vegetarian). She found that 40% of all participants exercise regularly, 30% are vegetarian, and 60% of vegetarians exercise regularly. How many non-vegetarian participants exercise occasionally?
Answer: 72 participants
Build the two-way table step by step: Total regular exercisers = 150 × 0.40 = 60. Total vegetarians = 150 × 0.30 = 45. Vegetarians who exercise regularly = 45 × 0.60 = 27. Therefore: vegetarians who exercise occasionally = 45 - 27 = 18, non-vegetarians total = 150 - 45 = 105, non-vegetarians who exercise regularly = 60 - 27 = 33, and non-vegetarians who exercise occasionally = 105 - 33 = 72. Choice A subtracts regular from total incorrectly. Choice C uses 90 - 27 instead of 105 - 33. Choice D adds occasional vegetarians to a wrong base.
- A survey of 500 customers examined their shopping method (online vs. in-store) and purchase amount (under $100 vs. $100+). The results showed that 180 customers made purchases under $100, 70 customers shopped online and spent under $100, and 290 customers shopped in-store. How many customers shopped online and spent $100 or more?
Answer: 140 customers
Total customers = 500. In-store shoppers = 290, so online shoppers = 500 - 290 = 210. Customers under $100 = 180, online shoppers under $100 = 70. Online shoppers $100+ = 210 - 70 = 140. Other choices represent common arithmetic errors.
- A company surveyed 200 employees about work location preference (Office or Remote) and department (Sales or Marketing). If 35% work in Sales, 45% prefer remote work, and the probability that a randomly selected employee works in Sales AND prefers remote work is 0.21, what is the marginal frequency for Marketing employees who prefer office work?
Answer: 58 employees
Set up the two-way table: Sales employees = 200 × 0.35 = 70, Marketing employees = 200 - 70 = 130, total preferring remote = 200 × 0.45 = 90, Sales employees preferring remote = 200 × 0.21 = 42. Therefore: Sales preferring office = 70 - 42 = 28, Marketing preferring remote = 90 - 42 = 48, Marketing preferring office = 130 - 48 = 58. Choice A incorrectly uses 130 - 42. Choice B uses total office workers (110). Choice D uses Sales preferring office plus an error.
- A school collected data on 180 students' transportation method (Bus or Car) and lunch location (Cafeteria or Outside). The data shows that 32 of students eat in the cafeteria, 95 take the bus, and 109 of bus riders eat in the cafeteria. What fraction of students who eat outside take the car?
Answer: 65
Build the two-way table: Students eating in cafeteria = 32×180=120, eating outside = 60, taking bus = 95×180=100, bus riders eating in cafeteria = 109×100=90. Therefore: bus riders eating outside = 100 - 90 = 10, car riders eating outside = 60 - 10 = 50, total eating outside = 60. Fraction of outside eaters taking car = 6050=65. Choice B would result from using 6040. Choice C from using 6048. Choice D from incorrect probability calculation.
