Mean, Median, and Mode - SHSAT Math

A student recorded the number of books read by classmates over summer break. When arranged in order, the data set is: 2, 3, 5, 7, 9, 12, 15. If two additional students report reading 6 books and 8 books respectively, what is the difference between the new median and the original mean, rounded to the nearest tenth?
Answer: $0.6$ Original mean: the sum is $2 + 3 + 5 + 7 + 9 + 12 + 15 = 53$ , so the mean is $\frac{53}{7} \approx 7.571$ . New median: add 6 and 8 and put the 9 values in order: 2, 3, 5, 6, 7, 8, 9, 12, 15. The middle (5th) value is $7$ . Difference: $7.571 - 7 = 0.571$ , which is $\frac{4}{7}$ . Rounded to the nearest tenth, this is $0.6$ . Comparing the original median with the new median gives $7 - 7 = 0$ , but the question compares the new median with the original mean. Using the new mean, $\frac{67}{9} \approx 7.444$ , instead of the original mean gives about $0.4$ . The difference between the new median and the original mean, rounded to the nearest tenth, is $0.6$ .
The scatter plot shown displays the number of hours studied and the test scores for 10 students. Based only on the data points shown, what is the median test score?
Answer: 80 Reading the 10 test scores from the scatter plot: 62, 68, 72, 76, 78, 82, 85, 88, 90, 94. With 10 values, the median is the average of the 5th and 6th: (78 + 82)/2 = 80. Answer (C). (A) and (B) are individual values (5th alone or 4th) rather than the average; (D) is the 6th value alone. Many students forget to average the two middle values when n is even.
In a biology class, test scores have a mean of 78 and a median of 82. If every student's score is increased by 6 points, what will be the new mean and median?
Answer: Mean = 84, Median = 88 (both measures increase by 6) When a constant is added to every value in a data set, both the mean and median increase by that same constant. Original mean = 78, so new mean = 78 + 6 = 84. Original median = 82, so new median = 82 + 6 = 88. This is a fundamental property of measures of central tendency under linear transformations. Choice A incorrectly assumes no change. Choices B and C incorrectly assume only one measure changes.
The numbers $3,5,8,9,10$ are written on the board. If another number is added so that the mean of all six numbers is 8, what number must be added?
Answer: 13 When you encounter a question about finding a missing value given the mean, you're working with the fundamental relationship between mean, sum, and count of data values. To find the missing number, start with what you know. The five given numbers are 3, 5, 8, 9, and 10. When the sixth number is added, all six numbers should have a mean of 8. Since mean equals total sum divided by count, you can write: $\frac{\text{sum of all 6 numbers}}{6} = 8$ This means the sum of all six numbers must equal $8 \times 6 = 48$ . The sum of the five given numbers is $3 + 5 + 8 + 9 + 10 = 35$ . Therefore, the missing sixth number must be $48 - 35 = 13$ . Let's verify: $\frac{3 + 5 + 8 + 9 + 10 + 13}{6} = \frac{48}{6} = 8$ ✓ Looking at the wrong answers: Choice B (12) would give a mean of $\frac{47}{6} = 7.83$ , which is too small. Choice C (14) would give $\frac{49}{6} = 8.17$ , slightly too large. Choice D (15) would give $\frac{50}{6} = 8.33$ , significantly too large. The correct answer is A. Strategy tip: For mean problems involving a missing value, always work backwards from the target mean. Multiply the desired mean by the total count to find the required sum, then subtract the known values to find what's missing. This systematic approach prevents calculation errors and works every time.
The mean of the numbers $7,\;9,\;x,\;11,\;13$ is 10. What is the value of $x$ ?
Answer: 10 When you encounter a problem involving the mean (average) of a set of numbers, remember that the mean equals the sum of all values divided by the count of values. Here, you need to work backwards from the given mean to find the missing value. Since the mean of the five numbers $7, 9, x, 11, 13$ is 10, you can set up the equation: $\frac{7 + 9 + x + 11 + 13}{5} = 10$ Multiply both sides by 5: $7 + 9 + x + 11 + 13 = 50$ Simplify the left side: $40 + x = 50$ Therefore: $x = 10$ Looking at the wrong answers: Choice A (8) would give a sum of 48, making the mean 9.6, not 10. Choice C (12) would create a sum of 52, resulting in a mean of 10.4. Choice D (14) would produce a sum of 54, giving a mean of 10.8. Each of these represents what you'd get if you made calculation errors or misunderstood the setup. Choice B (10) is correct because it's the only value that makes the total sum equal 50, which divided by 5 gives the required mean of 10. Strategy tip: When finding a missing value in a mean problem, always multiply the given mean by the total number of values first—this gives you the required sum. Then subtract the known values to find what's missing. This approach prevents arithmetic mistakes and works every time.
Based on the dot plot below, what is the mean number of pets per student?
Answer: 1.3 Compute using the frequencies: $\frac{0(2)+1(4)+2(3)+3(1)}{2+4+3+1}=\frac{13}{10}=1.3.$ Why the others are wrong: A. 1.2 is obtained by mis-counting one dot. C & D inflate the numerator by treating two-pet owners as three-pet owners.
For the four numbers $2,\,x,\,10,\,14$ the median is 9. What is the value of $x$ ?
Answer: 8 When you encounter a median problem with an unknown value, you need to understand that the median is the middle value when numbers are arranged in order. With four numbers, the median is the average of the two middle values. Let's organize the known numbers: $2, 10, 14$ . Since we need to find where $x$ fits, let's consider different cases. If $x$ is small (like 8), our ordered list would be $2, 8, 10, 14$ . The two middle values are 8 and 10, so the median would be $\frac{8 + 10}{2} = 9$ . This matches our target median of 9. Let's verify this is the only possibility. If $x = 9$ (choice B), our ordered list becomes $2, 9, 10, 14$ , giving us a median of $\frac{9 + 10}{2} = 9.5$ . If $x = 7$ (choice C), we get $2, 7, 10, 14$ with median $\frac{7 + 10}{2} = 8.5$ . If $x = 10$ (choice D), we have $2, 10, 10, 14$ with median $\frac{10 + 10}{2} = 10$ . Only choice A gives us the correct median of 9. Choice B creates a median of 9.5, choice C gives 8.5, and choice D results in 10. Each of these fails because they don't produce the required median. Strategy tip: Always arrange your numbers in order first, then remember that with an even number of values, you average the two middle numbers. Test your answer by calculating the actual median to confirm it matches the given condition.
The data set $2,4,6,8,10$ has a median of 6. If the value 100 is added, what is the new median?
Answer: 7 When you encounter median problems, remember that the median is the middle value when data is arranged in order. For sets with an even number of values, the median is the average of the two middle numbers. Let's work through this step by step. The original data set $2, 4, 6, 8, 10$ has 5 values, so the median is the 3rd value: 6. When we add 100, our new data set becomes $2, 4, 6, 8, 10, 100$ . Now we have 6 values (an even number), so we need the average of the 3rd and 4th values when arranged in order. The ordered set is still $2, 4, 6, 8, 10, 100$ . The 3rd value is 6 and the 4th value is 8. The median is $\frac{6 + 8}{2} = 7$ . Looking at the wrong answers: Choice A (6) assumes the median stays the same, which would only happen if we added a value that didn't shift the middle position. Choice C (8) incorrectly takes just the 4th value instead of averaging the middle two. Choice D (11) might result from incorrectly averaging 6 and 16, or some other calculation error involving the added value. Strategy tip: When the number of data points changes from odd to even (or vice versa), the method for finding the median changes too. Always recount your data points after adding or removing values, then determine whether you need a single middle value or the average of two middle values.
Which statement about the data set $1,1,2,3,3,4,4,5$ is true?
Answer: The modes are 1, 3, and 4. When you encounter a question about mode, you're looking for the value(s) that appear most frequently in a data set. The key is to count how many times each number appears and identify which ones tie for the highest frequency. Let's count the frequency of each value in the data set $1,1,2,3,3,4,4,5$ : - 1 appears 2 times - 2 appears 1 time - 3 appears 2 times - 4 appears 2 times - 5 appears 1 time The highest frequency is 2, and three different values (1, 3, and 4) all appear exactly 2 times each. Since these three values are tied for most frequent, they are all modes of this data set. Choice A is incorrect because the data set does have modes—multiple values appear with the same highest frequency. Choice B is wrong because while 1 does appear most frequently, it's not the only value that does so; 3 and 4 also appear 2 times each. Choice D makes the same error as B, identifying only one of the three modes instead of recognizing that multiple values tie for most frequent. Remember that a data set can have no mode (when all values appear equally), one mode (unimodal), two modes (bimodal), or even more modes (multimodal) when multiple values tie for the highest frequency. Always count carefully and identify all values that appear most frequently—don't stop after finding just one.
A data set consists of 12 values. When the values are arranged in ascending order, the 6th value is 45 and the 7th value is 51. If the mean of all 12 values is 48, what is the sum of all the values in the data set?
Answer: $576$ If the mean of 12 values is 48, then the sum of all values = 12 × 48 = 576. The information about the 6th and 7th values (which give us the median = (45+51)/2 = 48) is consistent with the mean but doesn't change the calculation for the sum. Choice A uses 11 instead of 12 values (11 × 48). Choice C incorrectly adds the median to the sum (576 + 12). Choice D uses the wrong mean (52 instead of 48).
The histogram shown groups the test scores of 40 students into five intervals. Using the midpoint of each interval to estimate scores, what is the approximate mean score?
Answer: 76.5 Use the midpoint of each interval as the score for every student in it, then take a weighted average. Midpoints and frequencies from the histogram: 55 (4 students), 65 (6), 75 (14), 85 (12), 95 (4). The total is $4 + 6 + 14 + 12 + 4 = 40$ students. Products: $55 \times 4 = 220$ , $65 \times 6 = 390$ , $75 \times 14 = 1050$ , $85 \times 12 = 1020$ , $95 \times 4 = 380$ . Their sum is 3060. Mean $= 3060 \div 40 = 76.5$ . The value 75.5 comes from a total that is 40 too small (3020 instead of 3060), an arithmetic slip in one of the products. The value 75 would be the plain average of the five midpoints, which ignores how many students fall in each interval. The values 74 and 72.5 give too much weight to the lower intervals.
For the data set $5,7,7,10,12$ , what is the value of (mean) $-$ (mode)?
Answer: 1.2 Statistics problems on the SHSAT often test your ability to calculate and compare different measures of central tendency. When you see a question asking for the difference between mean and mode, you need to find each value separately, then subtract. To find the mean of the data set $5, 7, 7, 10, 12$ , add all values and divide by the number of data points: $\frac{5 + 7 + 7 + 10 + 12}{5} = \frac{41}{5} = 8.2$ . The mode is the value that appears most frequently in the data set. Since 7 appears twice while all other values appear only once, the mode is 7. Therefore, (mean) - (mode) = $8.2 - 7 = 1.2$ . Looking at the wrong answers: Choice A) 1.0 likely results from incorrectly calculating the mean as 8 instead of 8.2, perhaps from an arithmetic error when adding the data values. Choice C) 1.3 could come from miscounting the sum as 42 instead of 41, giving a mean of 8.4. Choice D) 1.4 might result from calculating the mean as 8.4 and then making an additional subtraction error. Remember that the mode is always one of the actual data values in your set, while the mean can be any number. Double-check your arithmetic when finding the sum, and make sure you're dividing by the correct number of data points. Practice identifying the mode by looking for repeated values systematically.
The data set $15,\,20,\,25,\,20,\,30$ is given. Which measure or measures of central tendency are equal to $20$ ?
Answer: Median and Mode only When you encounter questions about measures of central tendency, you need to calculate the mean, median, and mode separately to see which ones equal the given value. Let's work through the data set $15, 20, 25, 20, 30$ systematically: Mean: Add all values and divide by the number of data points: $\frac{15 + 20 + 25 + 20 + 30}{5} = \frac{110}{5} = 22$ Median: Arrange the data in order: $15, 20, 20, 25, 30$ . The middle value (3rd position) is $20$ . Mode: The value that appears most frequently is $20$ (it appears twice, while all others appear once). So the median and mode both equal $20$ , while the mean equals $22$ . Looking at the answer choices: Choice A says "Median only" — this is wrong because the mode also equals $20$ . Choice B says "Mode only" — this is also wrong because the median equals $20$ as well. Choice D says "Mean, Median, and Mode" — this is incorrect because the mean equals $22$ , not $20$ . Choice C correctly identifies that both the median and mode equal $20$ . Strategy tip: Always calculate all three measures separately rather than assuming relationships between them. The mean can be "pulled" toward extreme values, while the median resists outliers, and the mode depends entirely on frequency. Don't let the presence of repeated values trick you into thinking all measures will be equal.
The dot plot shown displays the scores of 12 students on a quiz. If the teacher discovers that the two lowest scores were recorded incorrectly and each should be increased by 3 points, what is the new median score?
Answer: 7.5 Original scores (from dot plot): 3, 4, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10. After correction, 3→6 and 4→7. New sorted data: 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10. With 12 values the median is the average of the 6th and 7th: $(7+8)/2 = 7.5$ . (A) ignores the sort after correction; (C) takes only the 7th value; (D) averages 8 and 9.
The stem-and-leaf plot shown displays the heights (in cm) of 15 plants in a greenhouse. Based on the plot, what is the positive difference between the mean and the median of the data?
Answer: 0.4 Data from stems 2,3,4,5 with leaves: 24, 25, 28; 31, 33, 35, 37, 39; 42, 44, 46, 48; 51, 53, 55. That's 15 values. Sum = 24+25+28+31+33+35+37+39+42+44+46+48+51+53+55 = 591. Mean = 591/15 = 39.4. Median is the 8th value = 39. Difference = |39.4 - 39| = 0.4.
The numbers $1,2,2,3,4,4$ are written on index cards. Which single number should be added so that 4 becomes the unique mode of the data set?
Answer: 4 When you encounter mode problems, remember that the mode is the value that appears most frequently in a data set. For a mode to be "unique," it must appear more times than any other number. Let's analyze the current data set: $1, 2, 2, 3, 4, 4$ . Right now, both 2 and 4 appear twice each, while 1 and 3 appear once each. This means we have a tie between 2 and 4 for the mode, so neither is the unique mode. To make 4 the unique mode, we need 4 to appear more frequently than any other number. Since 2 currently appears twice (tied with 4), we need 4 to appear at least three times. Adding another 4 would give us three 4's, making it appear more than any other number. Let's check why the other options fail. Choice A (adding 1) would give us two 1's, but 2 and 4 would still be tied at two appearances each. Choice B (adding 2) would give us three 2's, making 2 the unique mode instead of 4. Choice C (adding 3) would give us two 3's, but we'd still have the tie between 2 and 4 at two appearances each. Only choice D works: adding 4 creates the frequency pattern where 4 appears three times while no other number appears more than twice. Strategy tip: In mode problems, always count current frequencies first, then determine what needs to change to achieve the desired result. Look for ties that need to be broken.
During one week, the daily high temperatures (in °F) were $68,70,72,66,74,68,70$ . What was the median high temperature for the week?
Answer: 70 When you encounter a median question, you're finding the middle value of a dataset when arranged in order. The median is less affected by extreme values than the mean, making it a useful measure of central tendency. To find the median, first arrange the temperatures in ascending order: $66, 68, 68, 70, 70, 72, 74$ . Since there are 7 values (an odd number), the median is simply the middle value - the 4th number in this ordered list. Counting from either end: $66, 68, 68, \boxed{70}, 70, 72, 74$ . The median is 70°F. Looking at the wrong answers: Choice A (68) appears twice in the dataset, making it the mode (most frequent value), but not the median. This is a common confusion between measures of central tendency. Choice B (69) might tempt you if you incorrectly average the two middle-adjacent values, but that's unnecessary when you have an odd number of data points. Choice D (71) doesn't correspond to any standard statistical measure for this dataset and likely results from a calculation error. Remember this key distinction: for an odd number of values, the median is always one of the actual data points (the exact middle one). For an even number of values, you'd average the two middle numbers. Also, always arrange your data in order first - the median of unordered data is meaningless. When you see median questions on the SHSAT, your first step should always be sorting the numbers.
A data set has 9 values with a median of 12. The values below the median are: 3, 7, 8, 11. The values above the median are: 15, 17, 20, 24. What is the difference between the mean and the median of this data set?
Answer: $1.0$ The complete data set is: 3, 7, 8, 11, 12, 15, 17, 20, 24. The median is 12 (5th value out of 9). The mean = (3+7+8+11+12+15+17+20+24)/9 = 117/9 = 13. The difference = 13 - 12 = 1. Choice B might result from a calculation error. Choice C assumes the mean is 14. Choice D assumes the mean is 14.3.
What is the median of the following six numbers? $2,\;9,\;5,\;12,\;7,\;14$
Answer: 8 When you encounter a median problem, remember that the median is the middle value when numbers are arranged in order. For an even number of values, it's the average of the two middle numbers. First, arrange the six numbers in ascending order: $2, 5, 7, 9, 12, 14$ Since there are six numbers (even), the median is the average of the 3rd and 4th values. The 3rd number is 7 and the 4th number is 9, so the median is $\frac{7 + 9}{2} = \frac{16}{2} = 8$ Let's examine why the other answers are incorrect: Choice A (7) represents only the 3rd value in the ordered list. This is a common mistake—students sometimes think the median for an even set is just one of the middle values rather than their average. Choice C (7.5) appears to be the result of incorrectly averaging 7 with 8 instead of 7 with 9. This could happen if you miscounted the position of the middle values or made an arithmetic error. Choice D (9) represents only the 4th value in the ordered list. Like choice A, this shows confusion about what to do when there's an even number of values. Study tip: Always start median problems by ordering the numbers first. For an odd number of values, pick the exact middle. For an even number of values, average the two middle numbers. Count carefully to identify which positions are "middle"—with 6 numbers, positions 3 and 4 are your targets.
To the nearest tenth, what is the mean of $2.5,\;3.0,\;3.5,\;4.0,\;4.0,\;4.5$ ?
Answer: 3.6 Finding the mean (or average) of a data set is one of the most fundamental statistics concepts you'll encounter. When you see a question asking for the mean, remember that you need to add all values and divide by the number of values. To find the mean of $2.5, 3.0, 3.5, 4.0, 4.0, 4.5$ , start by adding all the values: $2.5 + 3.0 + 3.5 + 4.0 + 4.0 + 4.5 = 21.5$ . Next, count how many values you have—there are 6 numbers in this data set. Finally, divide the sum by the count: $\frac{21.5}{6} = 3.583\overline{3}$ . Since the question asks for the answer to the nearest tenth, you round $3.583\overline{3}$ to $3.6$ . Looking at the wrong answers: Choice A (3.5) might tempt you if you mistakenly found the median instead of the mean, since 3.5 falls right in the middle when you arrange these numbers in order. Choice C (3.7) could result from a rounding error—perhaps rounding $3.583\overline{3}$ incorrectly or making an arithmetic mistake in your addition. Choice D (3.8) is too high and likely comes from calculation errors in either the sum or the division step. The correct answer is B (3.6). Study tip: Always double-check your arithmetic when calculating means, especially your addition. Also, make sure you're answering what the question asks for—mean, median, and mode are different concepts that students often confuse under pressure.

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