- The two-way table shows the relationship between grade level and participation in extracurricular activities. What is the probability that a randomly selected 10th grader participates in extracurricular activities?
Answer: The probability is 7545 for a 10th grader to participate
Among 10th graders, 45 participate in activities out of 75 total 10th graders. The probability is 45/75 = 3/5. Choice A uses total participants (120) instead of total 10th graders. Choice C uses non-participants (30) as numerator. Choice D uses the total student population (200) as denominator. - The multiple line graph tracks the daily temperature, humidity, and wind speed for one week. On which day was the difference between temperature and humidity the smallest?
Answer: Friday had the smallest difference between temperature and humidity
Calculating differences (Temperature - Humidity): Monday: 75-60=15, Tuesday: 78-55=23, Wednesday: 80-50=30, Thursday: 82-45=37, Friday: 85-82=3, Saturday: 88-75=13, Sunday: 90-70=20. Friday has the smallest difference at 3 units. Choice A (Monday) has difference of 15. Choice B (Wednesday) has difference of 30. Choice D (Sunday) has difference of 20.
- The double bar graph compares the number of books read by boys and girls in grades 6-8 during the summer reading program. What is the ratio of total books read by girls to total books read by boys across all three grades?
Answer: The ratio of girls' total to boys' total is 5:4
From the graph: Girls read 45 + 52 + 48 = 145 books total. Boys read 36 + 44 + 36 = 116 books total. The ratio 145:116 simplifies to 5:4 (since 145 ÷ 29 = 5 and 116 ÷ 29 = 4). Choice A (3:2) would represent 174:116. Choice B (4:3) would represent approximately 155:116. Choice D (7:5) would represent approximately 162:116.
- Based on the histogram shown, which statement about the data distribution is most accurate?
Answer: The mean is greater than the median due to positive skewness
Add the bar heights to get the total: 5 + 12 + 18 + 15 + 8 + 2 = 60 scores. The median is the average of the 30th and 31st scores. Counting up, 5 scores fall in 0-10 and 17 fall below 20, so the median is not in 10-20. Through 20-30 the running total is 35, so the 30th and 31st scores are both in 20-30, about 27 by interpolation.
Using bin midpoints, the mean is (5·5 + 12·15 + 18·25 + 15·35 + 8·45 + 2·55) ÷ 60 = 1,650 ÷ 60 = 27.5. The bars fall off more slowly to the right (15, 8, 2) than to the left (12, 5), so the data have a longer right tail. A right (positive) skew pulls the mean above the median, which matches 27.5 > about 27.
The claim that exactly 25% (15 scores) fall below 15 cannot be checked, because the 10-20 bar does not show where its 12 scores sit. Negative skew would need a long left tail, which this histogram does not have. The mean is greater than the median due to positive skewness.
- The box plot shown represents the distribution of daily temperatures (in °F) for a city during March. Based on this plot, approximately what percentage of days had temperatures between 45°F and 55°F?
Answer: 50% of the days had temperatures in this range
A box plot splits the data into four parts that each hold about 25% of the values. In this plot the box runs from the first quartile, 45°F, to the third quartile, 55°F, with the median at 50°F.
The days between Q1 and Q3 are the middle half of the data: about 25% from Q1 to the median plus about 25% from the median to Q3. So about 50% of the days had temperatures between 45°F and 55°F.
25% is the share between Q1 and the median only, or between the median and Q3. 75% would be the share from the minimum to Q3, or from Q1 to the maximum. A box plot does not show individual values, but its quartiles are enough to estimate this percent, so the question can be answered.
- The stacked bar chart shows the composition of three different trail mixes by ingredient type. Which trail mix has the highest proportion of nuts?
Answer: Mix B has the highest proportion of nuts at approximately 50%
Looking at the stacked bars: Mix A has nuts from 20% to 60% (40% nuts). Mix B has nuts from 15% to 65% (50% nuts). Mix C has nuts from 25% to 60% (35% nuts). Mix B has the highest proportion at 50%. Choice A incorrectly calculates Mix A's proportion. Choice C incorrectly identifies Mix C. Choice D incorrectly states equal proportions.
- The circle graph shows how a student spends their 24-hour day. Based on this data, how many hours does the student spend on activities other than sleeping and school?
Answer: The student spends 9 hours on other activities during the day
From the circle graph: Sleep = 37.5% of 24 hours = 9 hours, School = 25% of 24 hours = 6 hours. Other activities = 24 - 9 - 6 = 9 hours. Alternatively, other activities represent 37.5% (homework + recreation + meals), which equals 9 hours. Choice A represents 25% of 24 hours. Choice B represents about 29% of 24 hours. Choice C represents 33% of 24 hours.
- The stem-and-leaf plot shows test scores for Ms. Garcia's math class. What is the median score for the class?
Answer: The median test score is 79
From the stem-and-leaf plot, the scores in order are: 65, 67, 72, 75, 76, 78, 78, 79, 79, 82, 84, 85, 88, 91, 93. With 15 scores, the median is the 8th value, which is 79. Choice A (78) is the 6th and 7th values. Choice C (80) is not in the data set. Choice D (81) is also not in the data set.
- The frequency table shows the distribution of student heights in a physical education class. What percentage of students are 66 inches or taller?
Answer: Approximately 50% of students are 66 inches or taller
First find the total number of students: 4 + 7 + 6 + 8 + 6 + 3 = 34.
Next count the students who are 66 inches or taller. That includes the 66-67 row (8), the 68-69 row (6), and the 70+ row (3): 8 + 6 + 3 = 17 students.
The percentage is 17 ÷ 34 = 0.5, which is 50%. Exactly half the class is 66 inches or taller.
A value near 47% comes from dividing by the wrong total or slipping on the division; 17 out of 34 is exactly one half. About 53% would mean counting 18 students, for example by adding a student from the 64-65 row. About 43% would mean leaving out a few of the taller students. Check the total and the count separately before dividing. Approximately 50% of students are 66 inches or taller.
- The dot plot shows the number of pets owned by students in Mr. Thompson's class. What is the mode of this data set?
Answer: The mode is 2 pets because it appears most frequently
Counting the dots: 0 pets (3 dots), 1 pet (5 dots), 2 pets (7 dots), 3 pets (4 dots), 4 pets (2 dots), 5 pets (1 dot). The value 2 pets appears 7 times, which is more frequent than any other value. Choice A incorrectly identifies 1 pet (5 occurrences). Choice C incorrectly identifies 3 pets (4 occurrences). Choice D is incorrect since there is a clear single mode.
- The table shows the number of students in each grade who participated in different after-school activities. If a student is randomly selected from those who participate in Drama, what is the probability that the student is in 10th grade?
Answer: 258
From the table, 8 students in 10th grade participate in Drama, and the total number of students participating in Drama is 12 + 8 + 5 = 25. Therefore, the probability is 258. Choice A uses the total number of students (45) as the denominator instead of just Drama participants. Choice C uses only 9th and 10th grade Drama participants (20) as the denominator. Choice D uses the number of 10th graders in all activities as the numerator. - The back-to-back stem-and-leaf plot compares test scores for two different teaching methods. What can be concluded about the effectiveness of the two methods?
Answer: Method A produces higher scores with less variation
Method A scores: 68, 72, 74, 76, 78, 82, 84, 86, 88, 90, 92 (median = 82, range = 24). Method B scores: 55, 58, 62, 65, 68, 70, 72, 75, 78, 80, 85 (median = 70, range = 30). Method A has both a higher median score and smaller range, indicating higher performance with less variability. Choice B incorrectly reverses which method is superior. Choices C and D focus only on variability while ignoring the significant difference in score levels.
- The table shows the results of rolling two dice 50 times and recording the sum. Based on this experimental data, what is the experimental probability of rolling a sum of 7?
Answer: The experimental probability of rolling a sum of 7 is 508
From the frequency table, a sum of 7 occurred 8 times out of 50 rolls. Experimental probability = 8/50 = 4/25. Choice B uses an incorrect frequency count. Choice C gives the theoretical probability (1/6 ≈ 0.167), not experimental. Choice D gives the theoretical probability in unreduced form (6/36 = 1/6), not experimental. - The table shows survey results about favorite school subjects from students in three different grades. If a student who chose Science as their favorite is randomly selected, what is the probability that this student is in 8th grade?
Answer: The probability is 4518 that the Science student is in 8th grade
Total students who chose Science: 12 + 15 + 18 = 45. Students in 8th grade who chose Science: 18. Probability = 18/45 = 2/5. Choice A uses total students (75) instead of Science students. Choice C uses only 8th grade totals (25). Choice D uses total Science students as numerator instead of 8th grade Science students. - The scatter plot displays the relationship between hours of study time and test scores for 20 students. Which conclusion is best supported by the data?
Answer: There is a moderate positive relationship, but some students score well with minimal study time
The scatter plot shows a general upward trend (positive correlation) between study hours and test scores, but with notable scatter around the trend line. Some students achieve high scores (80+) with only 2-3 hours of study. Choice A is incorrect because one student studied 9 hours but scored only 78. Choice B is wrong because the correlation is positive, not negative. Choice D is incorrect because there is considerable variation around any potential trend line.
- The scatter plot shows the relationship between hours of TV watching per week and GPA for high school students. Which statement best describes what the data reveals about outliers?
Answer: One student watches 25 hours of TV but maintains a 3.8 GPA, making them an outlier
The general trend shows that more TV watching correlates with lower GPA. A student watching 25 hours (high TV time) but maintaining a 3.8 GPA (high academic performance) goes against this trend, making them an outlier. Choice A is incorrect because outliers are present. Choice C is incorrect because low GPA with low TV time follows the negative correlation. Choice D is incorrect because 30 hours/1.5 GPA follows the expected trend.
- The frequency polygon displays test score distributions for two different classes. Based on the graph, which statement about the two classes is most accurate?
Answer: Class B shows more variability in scores with a wider distribution range
Class B's frequency polygon shows data spread from 40 to 95 points with relatively even distribution, indicating high variability. Class A's data clusters more tightly around 70-85. Choice A is incorrect because Class A's peak is around 75-80, not 85. Choice C cannot be determined from frequency polygons alone. Choice D is incorrect because Class A shows one peak (unimodal), and Class B is not clearly normal.
- The cumulative frequency graph shows the distribution of test scores for 80 students. How many students scored between 70 and 85 points?
Answer: 30 students scored between 70 and 85 points
From the cumulative frequency graph: at score 85, cumulative frequency is 65; at score 70, cumulative frequency is 35. Students scoring between 70 and 85 = 65 - 35 = 30 students. Choice A (20) would be the difference between other points. Choice B (25) might result from misreading the graph values. Choice D (35) is the cumulative frequency at 70, not the difference.
- The histogram displays the distribution of quiz scores for a statistics class. Based on the shape of the distribution, which measure of central tendency would be most appropriate to report?
Answer: The median would be most appropriate because the distribution is skewed left
The histogram shows higher frequencies on the right side (higher scores) with a tail extending to the left (lower scores), indicating left skew. When data is skewed, the median is more appropriate than the mean because it's less affected by extreme values. Choice A is incorrect because the distribution is not symmetric. Choice C incorrectly identifies the skew direction. Choice D is incorrect because mode is rarely the most appropriate single measure of central tendency.
- The bar chart displays quarterly sales data for two products over one year. In which quarter was the difference between Product X and Product Y sales the greatest?
Answer: Quarter 4 had the greatest difference in sales between products
Calculating differences: Q1: |45-30| = 15, Q2: |38-42| = 4, Q3: |52-35| = 17, Q4: |60-25| = 35. Quarter 4 has the largest difference of 35 units. Choice A (Q1) has difference of 15. Choice B (Q2) has the smallest difference of 4. Choice C (Q3) has difference of 17.
