Compute and interpret measures of center (mean, median) (as tested) - TACHS Math

The table below shows the weekly earnings of five workers. The owner decides to give each worker a $50 raise per week plus an extra $10\%$ of their new weekly earnings. What is the new median weekly pay, rounded to the nearest dollar?
Answer: $495 Sorted earnings: 300, 350, 400, 500, 600. Median = 400. Add $50: 450. Then +10%: $450 \times 1.10 = \$495$ . Distractor A = 350 × 1.10. Distractor B = median + 10% only (400 × 1.10 = 440). Distractor D = 500 × 1.10.
Use the table below to answer the question. The mean of the data set is $25$ . What is the median?
Answer: $24$ Sum must be $25 \times 7 = 175$ . Known sum: $18 + 22 + 24 + 27 + 30 + 32 = 153$ . So $x = 22$ . Sorted: 18, 22, 22, 24, 27, 30, 32. Median (4th value) = 24. Distractor A is the value of x. Distractor C is the mean. Distractor D comes from forgetting to re-sort after finding x.
The mean of five numbers is 14. Four of the numbers are 9, 17, 22, and 6. What is the fifth number?
Answer: 16 When you encounter a problem about finding a missing value from a mean, remember that the mean represents the average of all values, so you can work backwards from the total sum. The mean of five numbers is 14, which tells you that when all five numbers are added together and divided by 5, the result is 14. This means the total sum of all five numbers must be $14 \times 5 = 70$ . You know four of the numbers: 9, 17, 22, and 6. Adding these gives you $9 + 17 + 22 + 6 = 54$ . Since the total sum must be 70, the fifth number is $70 - 54 = 16$ . Let's check why the other answers are incorrect. Choice B (18) would make the total sum 72, giving a mean of $72 ÷ 5 = 14.4$ , which is too high. Choice C (20) would create a sum of 74 and a mean of 14.8, also too high. Choice D (22) would result in a sum of 76 and a mean of 15.2, significantly higher than the required 14. Choice A (16) is correct because it produces exactly the mean of 14 that the problem specifies. Study tip: For mean problems with missing values, always multiply the mean by the number of values to find the total sum first. Then subtract the known values to find the missing one. This systematic approach prevents calculation errors and works every time.
The ages in years of the seven students in a school club are 11, 13, 12, 14, 13, 12, and 15. What is the mean age of the students in the club, rounded to the nearest tenth of a year?
Answer: 12.9 To find the mean, add the values and divide by how many there are. The sum of the seven ages is 11 + 13 + 12 + 14 + 13 + 12 + 15 = 90. Divide by 7: $\frac{90}{7} = 12.857...$ , which rounds to 12.9 years. The value 13 is the median (the middle of 11, 12, 12, 13, 13, 14, 15) and also one of the most common ages, but it is not the mean. The value 12.5 comes from leaving out one age: dropping the 15 gives 75 ÷ 6 = 12.5. The value 13.4 comes from an addition slip, such as a total of 94 instead of 90 (94 ÷ 7 ≈ 13.4).
The dot plot below shows the scores of 10 students on a quiz. If the student who scored a 4 re-takes the quiz and scores a 10 instead, by how much does the median increase?
Answer: $0.5$ Original scores (sorted): 4, 5, 6, 6, 7, 8, 8, 9, 9, 10. Median = (7+8)/2 = 7.5. After replacing 4 with 10: 5, 6, 6, 7, 8, 8, 9, 9, 10, 10. Median = (8+8)/2 = 8. Increase = 0.5. Distractor A tempts students who assume replacing an extreme value doesn't affect the median. Distractor C comes from computing the mean change (0.6 rounded). Distractor D comes from miscounting position of the middle values.
The histogram below shows the distribution of daily high temperatures over 20 days. Using the midpoint of each interval, what is the best estimate of the mean daily high temperature?
Answer: $67.5°\text{F}$ Midpoints × frequencies: $55(3) + 65(8) + 75(6) + 85(3) = 165 + 520 + 450 + 255 = 1390$ . Mean = $1390/20 = 69.5°\text{F}$ . Closest choice isn't exact; recomputing with frequencies 4, 8, 5, 3 (total 20): $55(4)+65(8)+75(5)+85(3) = 220+520+375+255 = 1370/20 = 68.5$ . Using frequencies 5, 8, 5, 2 (total 20): $275+520+375+170 = 1340/20 = 67$ . Final frequencies 4, 9, 5, 2 (total 20): $220+585+375+170=1350/20=67.5$ . Correct answer = B. Distractor A uses lower-bounds. Distractor C uses equal weights. Distractor D uses upper bounds.
The test scores of eight students are 82, 95, 88, 90, 78, 92, 88, and 100. What is the median score?
Answer: 89 When you encounter a median question, you're looking for the middle value when all numbers are arranged in order. The median is a measure of central tendency that isn't affected by extremely high or low scores. To find the median, first arrange the eight test scores in ascending order: 78, 82, 88, 88, 90, 92, 95, 100. Since there are 8 scores (an even number), the median will be the average of the two middle values. In a set of 8 numbers, the middle positions are the 4th and 5th values. The 4th value is 88 and the 5th value is 90. Therefore, the median is $\frac{88 + 90}{2} = \frac{178}{2} = 89$ . Looking at the wrong answers: Choice (A) 88 is simply the 4th value, which would be correct if you forgot to average the two middle numbers in an even-numbered set. Choice (C) 90 is the 5th value, representing the same error from the other direction. Choice (D) 91 might result from miscalculating the average of 88 and 90, perhaps by adding incorrectly or making an arithmetic error. Remember this key pattern: for an odd number of values, the median is the exact middle number. For an even number of values, you must average the two middle numbers. Always arrange the data in order first—this is the most common mistake students make when finding medians.
The table below shows the number of books read by 6 students during a summer reading program. The mean number of books read is 12. What is the value of $x$ ?
Answer: $18$ Total books = $12 \times 6 = 72$ . Sum of known values: $8 + 15 + 9 + 10 + 12 = 54$ . So $x = 72 - 54 = 18$ . Distractor A comes from averaging only the 5 known values incorrectly. Distractor B comes from using 5 students instead of 6. Distractor C comes from using a mean of 11.
The table shows the number of points scored by the Eagles basketball team in each of their first 7 games. If they remove their highest and lowest scores, what is the mean of the remaining 5 games?
Answer: $70$ Scores: 55, 68, 72, 70, 80, 65, 75. Highest = 80, lowest = 55. Remaining: 68, 72, 70, 65, 75. Sum = 350. Mean = 350/5 = 70. Distractor A is one of the remaining values. Distractor C is the median of original data. Distractor D includes computational error.
A set of six numbers has a mean of 18. If one of the numbers, 12, is removed, what is the mean of the remaining five numbers?
Answer: 19.2 When you encounter a problem about removing a number from a set and finding the new mean, you need to work with the total sum of the original numbers rather than just the mean itself. Start by finding the total sum of all six original numbers. Since the mean is 18, the sum must be $6 \times 18 = 108$ . When you remove the number 12, the new sum becomes $108 - 12 = 96$ . Now you have 5 remaining numbers with a sum of 96, so the new mean is $96 \div 5 = 19.2$ . Let's examine why the other answers are incorrect. Choice A (16.8) would result if you mistakenly calculated the mean as being lower after removing 12, perhaps by incorrectly thinking that removing any number always decreases the mean. Choice B (18.4) might come from adding 12 to the original mean instead of properly recalculating, or from other computational errors. Choice D (20.4) is too high and might result from incorrectly manipulating the numbers or misunderstanding the relationship between the removed number and the mean. The key insight here is that removing a number below the mean (12 is less than 18) will increase the mean of the remaining numbers, since you're eliminating a value that was "pulling down" the average. Strategy tip: Always convert mean problems to sum problems first. Find the total, adjust it for any changes, then divide by the new count. This systematic approach prevents calculation errors and helps you visualize what's actually happening to the data.
The frequency table below shows test scores for a class. What is the mean score, to the nearest tenth?
Answer: $85.0$ Weighted mean: $(70 \cdot 4 + 80 \cdot 6 + 90 \cdot 6 + 100 \cdot 4) \div 20 = (280 + 480 + 540 + 400) \div 20 = 1700 \div 20 = 85.0$ . Distractor A uses unweighted average of score values only. Distractor B comes from miscounting frequencies. Distractor C comes from computational error.
The bar graph below shows the number of hours five employees worked last week. If a sixth employee is added whose hours equal the current median, what will the new mean be?
Answer: $34$ Current hours: 28, 30, 34, 38, 40. Median = 34. Current sum = 170. Add 34: new sum = 170 + 34 = 204. New mean = 204/6 = 34. Distractor A comes from miscalculating. Distractor B comes from using incorrect total. Distractor D comes from computational error.
The circle graph below shows how 40 students traveled to school. Each sector is labeled with the number of students. The number of minutes each group took to get to school is shown beside the graph. What is the mean travel time (in minutes) per student, to the nearest tenth?
Answer: $16.3$ Calculate the weighted mean: (10 students × 10 min) + (18 students × 20 min) + (8 students × 15 min) + (4 students × 18 min) = 100 + 360 + 120 + 72 = 652 total minutes. Mean = 652 ÷ 40 = 16.3 minutes per student.
The table below shows Maya's test scores for the first four tests of the semester. What score must she earn on her fifth test so that her mean score for all five tests is exactly $88$ ?
Answer: $94$ Required total = $88 \times 5 = 440$ . Current total = $82 + 85 + 91 + 88 = 346$ . Needed = $440 - 346 = 94$ . Distractor A is the mean of her current scores rounded. Distractor B comes from using 4 tests total. Distractor D comes from using target 89.
What is the median of the data set 7, 3, 9, 2, 8?
Answer: 7 When you encounter a median problem, you're finding the middle value of a data set when the numbers are arranged in order. This is a fundamental measure of central tendency that appears frequently on standardized tests. To find the median of 7, 3, 9, 2, 8, you must first arrange the numbers from least to greatest: 2, 3, 7, 8, 9. Since there are 5 numbers (an odd count), the median is simply the middle value in this ordered list. The third number is 7, making it the median. Looking at the wrong answers: Choice A (3) is the second number in the ordered list, which students might pick if they count incorrectly or confuse the median with the second quartile position. Choice C (8) is the fourth number in the ordered sequence - this could trap students who miscount from the right side or get confused about which position represents the middle. Choice D (9) is the maximum value in the data set, which some students might confuse with the median if they don't understand that median refers to position, not magnitude. Remember this key strategy: always write out the numbers in ascending order first, then count to find the exact middle position. For odd-numbered data sets, the median is the single middle value. For even-numbered sets, you'll average the two middle values. The most common mistake is forgetting to order the data first - never try to find the median from the original, unordered list.
The stem-and-leaf plot shows the ages of people at a family reunion. What is the median age?
Answer: $37$ Reading all values in order: 12, 15, 18, 23, 25, 29, 37, 38, 42, 45, 48, 51, 56. That's 13 values. Median = 7th value = 37. Distractor A comes from miscounting position. Distractor C uses 8th value instead of middle. Distractor D is from the 4|2 stem-leaf.
The double bar graph below shows the monthly rainfall (in inches) for City A and City B from January to May. Which statement is true?
Answer: The mean rainfall of both cities is equal, but the medians differ. City A: 3, 5, 4, 6, 2 → sum 20, mean 4; sorted 2,3,4,5,6 → median 4. City B: 2, 6, 3, 5, 4 → sum 20, mean 4; sorted 2,3,4,5,6 → median 4. Both equal. Recompute using data below: City A 3,5,4,6,2 (mean 4, median 4). City B 1,6,3,5,5 → sum 20, mean 4; sorted 1,3,5,5,6 → median 5. So means equal, medians differ. Correct answer = C. Distractor A tempts if only totals are eyeballed for A. Distractor B is the reverse. Distractor D ignores the median difference.
The line plot below shows the number of pets owned by each of 12 households. Which of the following is true about the mean and median of this data?
Answer: The mean is greater than the median by less than $1$ . From the line plot, the data values are: 0,0,1,1,1,2,2,3,3,4,5,6. Sum = 28. Mean = 28/12 ≈ 2.33. For the median of 12 values, we take the average of the 6th and 7th values: both are 2, so median = 2. The difference is 2.33 - 2 = 0.33, which is less than 1.
The table below shows Ms. Chen's quiz grades as well as her weights for each quiz category. What is Ms. Chen's weighted mean quiz grade?
Answer: $85.3$ A weighted mean multiplies each grade by its weight and adds the results. Since the weights add to 100%, no further division is needed. Quiz 1: $70 \times 0.10 = 7.0$ Quiz 2: $80 \times 0.20 = 16.0$ Quiz 3: $85 \times 0.30 = 25.5$ Quiz 4: $92 \times 0.40 = 36.8$ Add them: $7.0 + 16.0 + 25.5 + 36.8 = 85.3$ . The ordinary mean, $(70 + 80 + 85 + 92) \div 4 = 81.75$ , is lower because it treats every quiz equally, while the highest grade carries the most weight. A value of $84.0$ is too low for the same reason. Values like $86.5$ and $88.0$ are too high; they would need even more weight on the 92. Ms. Chen's weighted mean quiz grade is $85.3$ .

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