The following practice questions are similar to what you'll find: in one, you have to figure out a salesman's traveling schedule based on a graph of his distance travelled each day, and in the other, you need to use a bar graph to calculate how often a certain number was rolled using a pair of dice.
Practice questions
- Chris lives in Omaha, Nebraska. He works in regional sales and has to travel throughout the week to meet with clients. The graph shows the distance that Chris's car is away from his home over a period of time in a given week. Which of the following situations best fits the information recorded in the graph?
A. Chris leaves one of his clients, drives to another client, stays there a few days, and then returns home. B. Chris leaves one of his clients, drives to another client, and then drives back home. C. Chris leaves one of his clients, drives back home, stays for a few days, and then drives to another client. D. Chris leaves home, drives to a client, stays there a few days, and then drives to another client. E. Chris leaves home, drives to a client, stays there a few days, and then drives home.
- A probability and statistics class conducted a die-rolling experiment. Each trial of the experiment consisted of rolling a single, fair, six-sided die six times and counting the number of 3s that resulted from each roll. The class recorded the results for 100 trials in the graph. In approximately what percent of trials did the class roll 3s more than two times?
A. 3% B. 5% C. 10% D. 15% E. 85%
Answers and explanations
- The correct answer is Choice (A).
The graph shows that at the beginning of the week Chris is 200 miles from home. Sunday, he travels farther away, presumably to another client, with whom he spends Monday and Tuesday. Wednesday he travels back to the point that is 0 miles from his home. In other words, he drives home. The answer that describes this schedule is Choice (A).
For Choice (D) or (E) to be right, the graph would have to start at 0. The graph for Choice (C) would have to camp out at 0 for a couple of days in the middle of data.
Choice (B) is meant to trap those who didn't include the days that Chris stayed over at the location of the second client.
- The correct answer is Choice (D).
To determine in what percent of the trials a 3 came up more than 2 times, you have to divide the number of trials with more than two rolls of 3 by the total number of trials, which is 100.
The first two bars on the left of the graph show the number of trials with two or fewer rolls of 3. The first bar shows 35 rolls and the second shows 50. 50 + 35 = 85, which means that 85 of the trials resulted in one or two rolls of 3. But the question doesn't ask for the number of trials with two or fewer rolls of 3, so don't pick Choice (E).
Because there were 85 trials of two or fewer rolls of 3, there must have been 15 trials where a 3 was rolled more than 2 times: 100 – 85 = 15. And 15 is 15 percent of 100.