Math 1 Quiz: Measures Of Center
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Measures Of CenterQuestion 1 of 14

A company tracks daily sales over 20 days. The mean daily sales is $2,400 and the median is $2,200. On day 21, the sales total is $3,800. What can be concluded about the updated 21-day mean compared to the updated 21-day median?

The 21-day mean will be greater than the 21-day median by a larger margin than before
The 21-day mean will be less than the 21-day median for the first time
The 21-day mean and median will be approximately equal, around $2,300
The 21-day median will exceed the 21-day mean due to the high day-21 value
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Math 1 Quiz

Math 1 Quiz: Measures Of Center

Practice Measures Of Center in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Measures Of Center, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A company tracks daily sales over 20 days. The mean daily sales is $2,400 and the median is $2,200. On day 21, the sales total is $3,800. What can be concluded about the updated 21-day mean compared to the updated 21-day median?

  1. The 21-day mean will be greater than the 21-day median by a larger margin than before (correct answer)
  2. The 21-day mean will be less than the 21-day median for the first time
  3. The 21-day mean and median will be approximately equal, around $2,300
  4. The 21-day median will exceed the 21-day mean due to the high day-21 value
Explanation: Original mean (2,400)>originalmedian(2,400) > original median (2,200), indicating a right-skewed distribution. New mean = (20 × $2,400 + $3,800)/21 = $48,000 + $3,800)/21 = $2,466.67. For the new median with 21 values, we need the 11th value when ordered. Since $3,800 is much larger than both the original mean and median, it will be placed at the high end of the ordered list, so the 11th value will be one of the original middle values, keeping the median close to $2,200. The gap between mean and median increases: originally $200, now approximately $267.

Question 2

The waiting times (in minutes) for 11 customers at a service center are recorded. When arranged in order, the times are: 2, 4, 6, 8, 9, 12, 15, 18, 22, 25, 43. If the service center's policy states that any wait time over 30 minutes should be recorded as exactly 30 minutes for reporting purposes, what happens to the mean and median of the reported data?

  1. The mean decreases and the median increases compared to the original data
  2. Both the mean and median decrease, but the mean decreases by a larger amount
  3. The mean decreases and the median remains unchanged compared to the original data (correct answer)
  4. Both measures remain exactly the same since only one value is affected
Explanation: Original data: 2, 4, 6, 8, 9, 12, 15, 18, 22, 25, 43. Original median = 12 (6th value out of 11). Original mean = 164/11 ≈ 14.91. After applying the policy, 43 becomes 30: 2, 4, 6, 8, 9, 12, 15, 18, 22, 25, 30. New median = 12 (still the 6th value, unchanged). New mean = 151/11 ≈ 13.73. The mean decreases because we reduced the largest value from 43 to 30, but the median stays the same because the 6th position value (12) is unaffected by this change.

Question 3

Two sections of a statistics course took the same exam. Section 1 (25 students) had a mean score of 82 and median score of 85. Section 2 (35 students) had a mean score of 78 and median score of 76. When the professor combines both sections to analyze the overall performance, which statement about the combined measures is most accurate?

  1. The combined mean will be exactly 80, and the combined median will be exactly 80.5
  2. The combined mean will be closer to 78 than to 82, and the combined median cannot be determined
  3. The combined mean will be exactly 79.67, and the combined median will be between 76 and 85 (correct answer)
  4. Both the combined mean and median will equal the simple average of the individual means and medians
Explanation: Combined mean = (25×82 + 35×78)/(25+35) = (2050 + 2730)/60 = 4780/60 = 79.67 (exact). For the combined median with 60 students total, we need the average of the 30th and 31st values when all scores are ordered. Without knowing the exact distributions, we cannot determine the exact combined median, but we know it must fall somewhere between the range of all values. Since Section 1 has median 85 and Section 2 has median 76, and we're combining the groups, the combined median will be between 76 and 85.

Question 4

A company's sales team has 12 members. Their monthly sales figures (in thousands of dollars) have a mean of 45 and a median of 42. The top performer sells 78 thousand dollars worth of goods. If this top performer leaves the company, and assuming all other values remain the same, what happens to the measures of center for the remaining 11 team members?

  1. The mean decreases to approximately 42 and becomes equal to the median
  2. The mean decreases to approximately 42 and the median decreases to approximately 40
  3. Both the mean and median decrease proportionally to approximately 40 each
  4. The mean decreases to approximately 42 and the median remains at 42 (correct answer)
Explanation: When you encounter questions about measures of center, remember that mean and median respond differently to data changes. The mean is sensitive to every value, while the median only depends on the middle position(s). Let's calculate what happens when the top performer leaves. Originally, with 12 members having a mean of 45, the total sales were 12×45=54012 \times 45 = 540 thousand dollars. After removing the top performer (78), the remaining 11 members have total sales of 540−78=462540 - 78 = 462 thousand dollars. The new mean is 462÷11≈42462 ÷ 11 ≈ 42. For the median with 12 people, it's the average of the 6th and 7th values when arranged in order, which equals 42. When we remove the highest value (78), we're left with 11 people, so the median becomes the 6th value in the ordered list. Since we only removed the top value, all the middle positions shift down by one spot, but the value that was originally in the 6th position remains the same. Therefore, the median stays at 42. Choice A incorrectly assumes the mean equals the median after the change. Choice B wrongly suggests the median decreases when removing the highest value. Choice C incorrectly assumes both measures change proportionally, which isn't how these statistics behave. Study tip: Remember that removing extreme values (especially the highest or lowest) typically affects the mean more dramatically than the median. The median is resistant to outliers, while the mean is pulled toward extreme values.

Question 5

A quality control inspector measures the weights (in grams) of 7 manufactured parts: 98, 102, 99, 105, 101, 97, 103. Due to a calibration error, each measurement was 2 grams too high. After correcting all measurements by subtracting 2 grams from each, what is the relationship between the original and corrected measures of center?

  1. Both the mean and median of the corrected data are exactly 2 grams lower than the originals (correct answer)
  2. The mean decreases by 2 grams, but the median decreases by a different amount
  3. The median decreases by exactly 2 grams, but the mean decreases by 14 grams total
  4. Both measures decrease, but by amounts that depend on the distribution shape
Explanation: When every value in a dataset is decreased by the same constant amount (2 grams), both the mean and median decrease by exactly that same amount. This is because: (1) Mean: if each of n values decreases by 2, the sum decreases by 2n, so the mean decreases by 2n/n = 2. (2) Median: subtracting 2 from every value shifts the entire ordered list down by 2, so the middle value(s) also decrease by exactly 2. This is a fundamental property of measures of center under linear transformations.

Question 6

A fitness tracking app records the daily step counts for a user over two weeks (14 days). The data shows significant variation, with some days having very low step counts due to illness and some days having very high counts due to hiking activities.

The mean daily step count is 8,750 steps and the median is 7,200 steps. If the user wants to set a realistic daily goal that represents 'typical' performance while accounting for the fact that there are several unusually high days in the data, which measure of center should be used and why?

  1. Use the mean because it incorporates all data points and provides the most mathematically precise measure
  2. Use the median because it better represents typical performance when extreme values are present in the dataset (correct answer)
  3. Use the average of the mean and median (7,975 steps) to balance both measures appropriately
  4. Use the mean because the presence of low outliers makes it more conservative than the median
Explanation: Since mean (8,750) > median (7,200), the distribution is right-skewed, indicating some unusually high values (hiking days) are pulling the mean upward. The median is less affected by these extreme values and better represents the 'typical' day. For goal-setting purposes, the median provides a more achievable target that represents what the user actually does on a typical day, rather than an inflated average that includes exceptional hiking days.

Question 7

A teacher recorded the number of books read by students in two different classes over the summer. Class A has 12 students, and Class B has 18 students. The combined median for all 30 students is 7 books.

If Class A has a median of 6 books and Class B has a median of 8 books, what must be true about the distribution of the combined data?

  1. The 15th and 16th students in the combined ordered list both read exactly 7 books
  2. More than half of the students in the combined classes read fewer than 7 books
  3. The 15th student read 6 books and the 16th student read 8 books in the combined list (correct answer)
  4. At least 6 students from Class A and at least 9 students from Class B read 7 or more books
Explanation: With 30 students total, the median is the average of the 15th and 16th values when all students are ordered by books read. Given that Class A (12 students) has median 6 and Class B (18 students) has median 8, and the combined median is 7, we need (15th value + 16th value)/2 = 7, so 15th + 16th = 14. Since we're combining two groups where one has a lower median (6) and one has a higher median (8), and we need an average of 7 for the middle two values, the most logical scenario is that the 15th student read 6 books (likely from the higher end of Class A) and the 16th student read 8 books (likely from the lower end of Class B).

Question 8

A researcher collects data on the ages of participants in a study. The mean age is 34.5 years and the median age is 32 years. If the researcher realizes that one participant's age was incorrectly recorded as 28 when it should have been 58, how will correcting this error affect the measures of center?

  1. The mean will increase by exactly 30 years, and the median will also increase significantly
  2. The mean will increase by 30/n years (where n is sample size), and the median may increase slightly (correct answer)
  3. Both the mean and median will increase by exactly the same amount proportionally
  4. The median will increase more than the mean since 58 is much larger than 28
Explanation: When changing one value from 28 to 58 (an increase of 30), the mean increases by 30/n where n is the total number of participants. The median may increase slightly, remain the same, or increase moderately depending on where these values fall relative to the middle position(s), but the change will be much smaller than the change in the mean. Since 58 > mean (34.5) and 28 < mean, this correction moves a below-mean value to an above-mean value, so the mean definitely increases by 30/n.

Question 9

A researcher studying reaction times collects data from 20 participants. The mean reaction time is 0.45 seconds with a median of 0.42 seconds. Upon review, the researcher discovers that the fastest reaction time of 0.18 seconds was actually a measurement error and should be removed from the dataset. What can be concluded about the measures of center for the remaining 19 participants?

  1. Both the mean and median will increase, with the mean increasing by a larger amount
  2. The mean will increase while the median will either stay the same or increase slightly (correct answer)
  3. Both measures will decrease since we're removing data from the lower end of the distribution
  4. The median will increase significantly while the mean will change very little
Explanation: Since mean (0.45) > median (0.42), the distribution is right-skewed. Removing the smallest value (0.18) will increase the mean because we're removing a below-average value. New mean = (20 × 0.45 - 0.18)/19 = (9.0 - 0.18)/19 = 8.82/19 ≈ 0.464. For the median: with 20 values, the original median was the average of the 10th and 11th values (0.42). After removing the smallest value, we have 19 values, so the new median is the 10th value. Since we removed the 1st value from the ordered list, the new 10th value is the same as the original 11th value. The median will stay the same or increase slightly.

Question 10

A small bookstore tracks the number of books sold each day for two weeks. The data shows: Week 1: 12, 8, 15, 22, 18, 10, 25. Week 2: 14, 11, 19, 16, 13, 21, 20.

The owner wants to compare the two weeks using measures of center. Which comparison is most accurate?

  1. Week 1 has both higher mean and higher median than Week 2
  2. Week 2 has both higher mean and higher median than Week 1 (correct answer)
  3. Week 1 has higher mean but Week 2 has higher median
  4. Week 2 has higher mean but Week 1 has higher median
Explanation: Week 1: Mean = 110/7 ≈ 15.71, Median = 15 (sorted: 8,10,12,15,18,22,25). Week 2: Mean = 114/7 ≈ 16.29, Median = 16 (sorted: 11,13,14,16,19,20,21). Week 2 has both higher mean (16.29 vs 15.71) and higher median (16 vs 15). Students might miscalculate or confuse the weeks.

Question 11

A basketball coach recorded the points scored by her team in their last 12 games: 58, 62, 45, 71, 58, 49, 67, 58, 53, 61, 44, 56. If the team scores 70 points in their next game, which statement about the measures of center is correct?

  1. The mean will increase by more than the median will increase (correct answer)
  2. The median will increase by more than the mean will increase
  3. The mean and median will increase by the same amount
  4. The median will increase but the mean will decrease
Explanation: Original data sorted: 44, 45, 49, 53, 56, 58, 58, 58, 61, 62, 67, 71. Original mean = 682/12 ≈ 56.83, original median = (58+58)/2 = 58. With 70 added, new mean = 752/13 ≈ 57.85, new median = 58 (7th value in sorted list). Mean increases by about 1.02, median increases by 0. Choice A is correct.

Question 12

In a class of 25 students, the mean test score is 78. The teacher discovers that one score was incorrectly recorded as 65 when it should have been 85. After correcting this error, what is the difference between the new mean and the original mean?

  1. 0.8 points higher (correct answer)
  2. 20 points higher
  3. 1.25 points higher
  4. The mean remains unchanged
Explanation: The correction changes one score by +20 points (from 65 to 85). This adds 20 to the total sum. The new mean increases by 20/25 = 0.8 points. Choice B incorrectly uses the full 20-point change. Choice C might result from calculation errors. Choice D incorrectly assumes no change occurs.

Question 13

A company's quarterly profits for the past three years were: Q1: $45K, $52K, $48K; Q2: $38K, $41K, $44K; Q3: $55K, $61K, $58K; Q4: $42K, $49K, $46K (listed chronologically by year). The company wants to identify which quarter typically performs best. What should they conclude?

  1. Q3 performs best with mean $58K and median $58K (correct answer)
  2. Q1 performs best with mean $48.33K and median $48K
  3. Q4 performs best with mean $45.67K and median $46K
  4. Q2 performs best with mean $41K and median $41K
Explanation: Q1: Mean = (45+52+48)/3 = 48.33K, Median = 48K. Q2: Mean = (38+41+44)/3 = 41K, Median = 41K. Q3: Mean = (55+61+58)/3 = 58K, Median = 58K. Q4: Mean = (42+49+46)/3 = 45.67K, Median = 46K. Q3 has both the highest mean and highest median, making it the best performing quarter. Students might miscalculate means or confuse which quarter performs best.

Question 14

In a neighborhood, the median household income is $65,000 and the mean household income is $73,000. A new family moves in with an income of exactly $65,000. How do the measures of center change?

  1. The mean decreases and the median remains the same or changes minimally (correct answer)
  2. The mean increases and the median increases to a value above $65,000
  3. Both the mean and median decrease by the same proportional amount
  4. The mean remains unchanged and the median increases significantly above $65,000
Explanation: Since the original mean (73,000)isgreaterthantheoriginalmedian(73,000) is greater than the original median (65,000), the distribution is right-skewed. Adding a household with income equal to the current median ($65,000) will pull the mean down toward this value, since $65,000 < $73,000. The median will either stay exactly the same (if the dataset size goes from even to odd) or change very slightly, since we're adding a value at the current median.