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    Medium Mean, Median, Mode Practice Questions

    April 5, 202617 min read0 views
    Medium Mean, Median, Mode Practice Questions

    Medium Mean, Median, Mode Practice Questions

    Understanding the core measures of central tendency is a fundamental skill in statistics and data analysis. Being able to correctly calculate and interpret the mean, median, and mode allows you to summarize entire datasets with a single representative value. This guide provides a clear explanation of these concepts, worked-out examples, and a series of medium-difficulty practice questions to help you master the calculation and application of mean, median, and mode.

    Concept Explanation

    Mean, median, and mode are the three main measures of central tendency used in statistics to describe the center point or typical value of a dataset. Each measure provides a different perspective on the data's central value and is useful in different situations. Understanding how to calculate each one is crucial for introductory data analysis.

    What is the Mean?

    The mean is the most common measure of central tendency, often referred to as the "average." It is calculated by summing all the values in a dataset and then dividing by the total number of values.

    Formula: Mean = (Sum of all values) / (Number of values)

    The mean is sensitive to outliers, which are values that are significantly higher or lower than the rest of the data. A single outlier can dramatically shift the mean, which is why it's not always the best measure for skewed datasets.

    What is the Median?

    The median is the middle value in a dataset that has been arranged in ascending or descending order. It effectively splits the dataset into two equal halves. Unlike the mean, the median is robust to outliers, making it a better measure for skewed data.

    • For an odd number of values: The median is the single middle value.
    • For an even number of values: The median is the mean of the two middle values.

    To learn more about how different data distributions affect these measures, you can explore concepts like Z-scores, which quantify how far a data point is from the mean.

    What is the Mode?

    The mode is the value that appears most frequently in a dataset. A dataset can have one mode (unimodal), two modes (bimodal), more than two modes (multimodal), or no mode at all if every value appears only once. The mode is particularly useful for categorical data (e.g., favorite colors, types of cars) but can be used with numerical data as well.

    Solved Examples of Mean, Median, and Mode

    The following examples demonstrate how to calculate the mean, median, and mode for different types of datasets. Working through these will help solidify your understanding before you tackle the practice questions.

    Example 1: Basic Dataset (Odd Number of Values)

    Find the mean, median, and mode for the following dataset: {3, 7, 5, 13, 20, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29}

    1. Order the data: First, arrange the dataset in ascending order to easily find the median and mode.
      {3, 5, 7, 12, 13, 14, 20, 23, 23, 23, 23, 29, 39, 40, 56}
    2. Calculate the Mean: Sum all the numbers and divide by the count of numbers (15).
      Sum = 3 + 5 + 7 + 12 + 13 + 14 + 20 + 23 + 23 + 23 + 23 + 29 + 39 + 40 + 56 = 330
      Mean = 330 / 15 = 22
    3. Find the Median: The dataset has 15 values (an odd number). The median is the middle value, which is the (15+1)/2 = 8th value.
      The 8th value in the ordered list is 23.
      Median = 23
    4. Find the Mode: Look for the number that appears most frequently.
      The number 23 appears 4 times, which is more than any other number.
      Mode = 23

    Example 2: Dataset with an Outlier

    Consider the weekly salaries of 5 employees: { 500 , 500, 520, 550 , 550, 600, 2500}. Calculate the mean and median and comment on the difference.</strong></p> <ol> <li><strong>The dataset is already ordered:</strong> {500, 520 , 520, 550, 600 , 600, 2500}

  1. Calculate the Mean: Sum the salaries and divide by 5.
    Sum = 500 + 520 + 550 + 600 + 2500 = 4670 < b r > M e a n = 4670 <br>Mean = 4670 / 5 = 934 < / l i > < l i > < s t r o n g > F i n d t h e M e d i a n : < / s t r o n g > T h e d a t a s e t h a s 5 v a l u e s . T h e m e d i a n i s t h e m i d d l e ( 3 r d ) v a l u e . < b r > M e d i a n = 934</li> <li><strong>Find the Median:</strong> The dataset has 5 values. The median is the middle (3rd) value.<br> Median = 550
  2. Analysis: The mean ( 934 ) i s s i g n i f i c a n t l y h i g h e r t h a n t h e m e d i a n ( 934) is significantly higher than the median ( 550). This is because the outlier salary of 2500 p u l l s t h e a v e r a g e u p . I n t h i s c a s e , t h e m e d i a n i s a m o r e a c c u r a t e r e p r e s e n t a t i o n o f a " t y p i c a l " e m p l o y e e ′ s s a l a r y . T h i s i l l u s t r a t e s w h y u n d e r s t a n d i n g m e a s u r e s o f s p r e a d , l i k e < a h r e f = " / b l o g / v a r i a n c e − c a l c u l a t i o n − p r a c t i c e − q u e s t i o n s " > v a r i a n c e < / a > , i s a l s o i m p o r t a n t . < / l i > < / o l > < h 3 > E x a m p l e 3 : F i n d i n g a M i s s i n g V a l u e G i v e n t h e M e a n < / h 3 > < p > < s t r o n g > A s t u d e n t h a s s c o r e s o f 85 , 92 , 88 , a n d 78 o n f o u r q u i z z e s . W h a t s c o r e m u s t t h e s t u d e n t g e t o n t h e f i f t h q u i z t o h a v e a m e a n s c o r e o f 87 ? < / s t r o n g > < / p > < o l > < l i > < s t r o n g > S e t u p t h e m e a n f o r m u l a : < / s t r o n g > W e k n o w t h e d e s i r e d m e a n ( 87 ) a n d t h e n u m b e r o f v a l u e s ( 5 ) . L e t ′ x ′ b e t h e s c o r e o n t h e f i f t h q u i z . < b r > M e a n = ( S u m o f s c o r e s ) / ( N u m b e r o f s c o r e s ) < / l i > < l i > < s t r o n g > P l u g i n t h e k n o w n v a l u e s : < / s t r o n g > < b r > 87 = ( 85 + 92 + 88 + 78 + x ) / 5 < / l i > < l i > < s t r o n g > S o l v e f o r x : < / s t r o n g > < b r > M u l t i p l y b o t h s i d e s b y 5 : 87 ∗ 5 = 435 < b r > S u m t h e k n o w n s c o r e s : 85 + 92 + 88 + 78 = 343 < b r > S o , 435 = 343 + x < / l i > < l i > < s t r o n g > I s o l a t e x : < / s t r o n g > < b r > x = 435 − 343 = 92 < b r > T h e s t u d e n t m u s t s c o r e a 92 o n t h e f i f t h q u i z . < / l i > < / o l > < h 3 > E x a m p l e 4 : E v e n N u m b e r o f V a l u e s a n d a B i m o d a l D a t a s e t < / h 3 > < p > < s t r o n g > F i n d t h e m e a n , m e d i a n , a n d m o d e f o r t h e d a t a s e t o f d a i l y h i g h t e m p e r a t u r e s : 68 , 72 , 72 , 75 , 78 , 81 , 81 , 85 < / s t r o n g > < / p > < o l > < l i > < s t r o n g > T h e d a t a s e t i s a l r e a d y o r d e r e d : < / s t r o n g > 68 , 72 , 72 , 75 , 78 , 81 , 81 , 85 < / l i > < l i > < s t r o n g > C a l c u l a t e t h e M e a n : < / s t r o n g > S u m t h e v a l u e s a n d d i v i d e b y 8. < b r > S u m = 68 + 72 + 72 + 75 + 78 + 81 + 81 + 85 = 612 < b r > M e a n = 612 / 8 = 76.5 < / l i > < l i > < s t r o n g > F i n d t h e M e d i a n : < / s t r o n g > T h e d a t a s e t h a s 8 v a l u e s ( a n e v e n n u m b e r ) . T h e m e d i a n i s t h e a v e r a g e o f t h e t w o m i d d l e v a l u e s ( t h e 4 t h a n d 5 t h v a l u e s ) . < b r > T h e m i d d l e v a l u e s a r e 75 a n d 78. < b r > M e d i a n = ( 75 + 78 ) / 2 = 153 / 2 = 76.5 < / l i > < l i > < s t r o n g > F i n d t h e M o d e : < / s t r o n g > L o o k f o r t h e m o s t f r e q u e n t v a l u e ( s ) . < b r > B o t h 72 a n d 81 a p p e a r t w i c e . S i n c e t h e y a p p e a r w i t h t h e s a m e h i g h e s t f r e q u e n c y , t h e d a t a s e t i s b i m o d a l . < b r > M o d e s = 72 a n d 81 < / l i > < / o l > < h 2 > P r a c t i c e Q u e s t i o n s < / h 2 > < p > T e s t y o u r k n o w l e d g e w i t h t h e s e m e d i u m − l e v e l m e a n , m e d i a n , a n d m o d e q u e s t i o n s . D e t a i l e d s o l u t i o n s a r e p r o v i d e d b e l o w . < / p > < p > < s t r o n g > 1. < / s t r o n g > T h e n u m b e r o f g o a l s s c o r e d b y a s o c c e r t e a m i n t h e i r l a s t 11 m a t c h e s w e r e : 2 , 3 , 0 , 1 , 1 , 2 , 4 , 5 , 1 , 0 , 2. F i n d t h e m e a n , m e d i a n , a n d m o d e f o r t h e n u m b e r o f g o a l s s c o r e d . < / p > < p > < s t r o n g > 2. < / s t r o n g > A c o m p a n y ′ s m o n t h l y p r o f i t s f o r t h e f i r s t s i x m o n t h s o f t h e y e a r w e r e : 2500 pulls the average up. In this case, the median is a more accurate representation of a "typical" employee's salary. This illustrates why understanding measures of spread, like <a href="/blog/variance-calculation-practice-questions">variance</a>, is also important.</li> </ol> <h3>Example 3: Finding a Missing Value Given the Mean</h3> <p><strong>A student has scores of 85, 92, 88, and 78 on four quizzes. What score must the student get on the fifth quiz to have a mean score of 87?</strong></p> <ol> <li><strong>Set up the mean formula:</strong> We know the desired mean (87) and the number of values (5). Let 'x' be the score on the fifth quiz. <br>Mean = (Sum of scores) / (Number of scores)</li> <li><strong>Plug in the known values:</strong><br> 87 = (85 + 92 + 88 + 78 + x) / 5</li> <li><strong>Solve for x:</strong><br> Multiply both sides by 5: 87 * 5 = 435<br> Sum the known scores: 85 + 92 + 88 + 78 = 343<br> So, 435 = 343 + x</li> <li><strong>Isolate x:</strong><br> x = 435 - 343 = 92<br> The student must score a 92 on the fifth quiz.</li> </ol> <h3>Example 4: Even Number of Values and a Bimodal Dataset</h3> <p><strong>Find the mean, median, and mode for the dataset of daily high temperatures: {68, 72, 72, 75, 78, 81, 81, 85}</strong></p> <ol> <li><strong>The dataset is already ordered:</strong> {68, 72, 72, 75, 78, 81, 81, 85}</li> <li><strong>Calculate the Mean:</strong> Sum the values and divide by 8.<br> Sum = 68 + 72 + 72 + 75 + 78 + 81 + 81 + 85 = 612 <br>Mean = 612 / 8 = 76.5</li> <li><strong>Find the Median:</strong> The dataset has 8 values (an even number). The median is the average of the two middle values (the 4th and 5th values).<br> The middle values are 75 and 78. <br>Median = (75 + 78) / 2 = 153 / 2 = 76.5</li> <li><strong>Find the Mode:</strong> Look for the most frequent value(s).<br> Both 72 and 81 appear twice. Since they appear with the same highest frequency, the dataset is bimodal. <br>Modes = 72 and 81</li> </ol> <h2>Practice Questions</h2> <p>Test your knowledge with these medium-level mean, median, and mode questions. Detailed solutions are provided below.</p> <p><strong>1.</strong> The number of goals scored by a soccer team in their last 11 matches were: 2, 3, 0, 1, 1, 2, 4, 5, 1, 0, 2. Find the mean, median, and mode for the number of goals scored.</p> <p><strong>2.</strong> A company's monthly profits for the first six months of the year were: 12,000, 15 , 000 , 15,000, 11,500, 18 , 000 , 18,000, 14,000, and 16,500. Calculate the median monthly profit.</p> <p><strong>3.</strong> The mean age of a group of 8 friends is 24. If a new friend with an age of 33 joins the group, what will be the new mean age of the group?</p> <div class="bevinzey-product-cta" style="background: linear-gradient(135deg, #3b82f6, #8b5cf6); border-radius: 12px; padding: 32px; text-align: center; margin: 32px 0;"> <h3 style="color: #ffffff; font-size: 20px; margin-bottom: 8px;">Want unlimited practice questions like these?</h3> <p style="color: #e2e8f0; font-size: 15px; margin-bottom: 16px;">Generate AI-powered questions with step-by-step solutions on any topic.</p> <a href="/question-generator" style="display: inline-block; background: #ffffff; color: #3b82f6; padding: 12px 28px; border-radius: 8px; text-decoration: none; font-weight: 600; font-size: 15px;">Try Question Generator Free →</a> </div> <p><strong>4.</strong> The following dataset represents the number of daily visitors to a small museum: {15, 22, 18, 35, 19, 22, 11, 40, 22, 19}. Is this dataset unimodal, bimodal, or multimodal? What is the mode(s)?</p> <p><strong>5.</strong> The mean of five numbers is 30. Four of the numbers are 25, 35, 28, and 40. What is the fifth number?</p> <p><strong>6.</strong> Consider the dataset of test scores: {55, 65, 70, 72, 75, 80, 82, 95, 100}. If the lowest score (55) is removed from the dataset, which measure of central tendency (mean or median) will change more? Calculate both new values to verify.</p> <p><strong>7.</strong> A real estate agent sold 7 houses with the following prices:250,000, 275 , 000 , 275,000, 310,000, 295 , 000 , 295,000, 450,000, 310 , 000 , 310,000, 1,200,000. Which measure of central tendency, mean or median, would be a more appropriate representation of a typical house price in this set, and why?

    8. The mean weight of 10 basketball players is 200 lbs. If the heaviest player (240 lbs) and the lightest player (160 lbs) are removed, what is the new mean weight of the remaining 8 players?

    9. You have the following dataset: {10, 15, 20, 25, 30}. If you add the number 100 to this dataset, how does the median change compared to the mean?

    10. The mode of the following dataset is 18. What is the smallest possible integer value for x?
    {12, 18, 11, 25, 18, 12, x, 25}

    Answers & Explanations

    Here are the detailed solutions for the practice questions above.

    1. The number of goals scored by a soccer team in their last 11 matches were: 2, 3, 0, 1, 1, 2, 4, 5, 1, 0, 2. Find the mean, median, and mode.

    1. Order the data: {0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 5}
    2. Mean: Sum = 0+0+1+1+1+2+2+2+3+4+5 = 21. Count = 11.
      Mean = 21 / 11 ≈ 1.91
    3. Median: There are 11 (odd) values. The median is the (11+1)/2 = 6th value.
      The 6th value is 2. Median = 2.
    4. Mode: The value '1' appears 3 times, and '2' appears 3 times. '0' appears twice. Since '1' and '2' both appear with the highest frequency, this dataset is bimodal.
      Mode = 1 and 2.

    2. A company's monthly profits were: 12 , 000 , 12,000, 15,000, 11 , 500 , 11,500, 18,000, 14 , 000 , a n d 14,000, and 16,500. Calculate the median monthly profit.

    1. Order the data: { 11 , 500 , 11,500, 12,000, 14 , 000 , 14,000, 15,000, 16 , 500 , 16,500, 18,000}
    2. Find the Median: There are 6 (even) values. The median is the average of the two middle values (the 3rd and 4th).
      The middle values are 14 , 000 a n d 14,000 and 15,000.
      Median = ( 14 , 000 + 14,000 + 15,000) / 2 = 29 , 000 / 2 = 29,000 / 2 = 14,500.

    3. The mean age of a group of 8 friends is 24. If a new friend with an age of 33 joins the group, what will be the new mean age?

    1. Find the total current age: If the mean of 8 friends is 24, the sum of their ages is Mean × Count = 24 × 8 = 192.
    2. Add the new friend's age: The new total age is 192 + 33 = 225.
    3. Calculate the new mean: The group now has 9 people.
      New Mean = New Total Age / New Count = 225 / 9 = 25.

    4. Is the dataset {15, 22, 18, 35, 19, 22, 11, 40, 22, 19} unimodal, bimodal, or multimodal? What is the mode(s)?

    1. Count the frequency of each number:
      11: 1 time
      15: 1 time
      18: 1 time
      19: 2 times
      22: 3 times
      35: 1 time
      40: 1 time
    2. Identify the mode: The number 22 appears 3 times, which is more than any other number. Since there is only one mode, the dataset is unimodal.
    3. Answer: The dataset is unimodal, and the mode is 22.

    5. The mean of five numbers is 30. Four of the numbers are 25, 35, 28, and 40. What is the fifth number?

    1. Find the required sum: If the mean of 5 numbers is 30, their sum must be Mean × Count = 30 × 5 = 150.
    2. Sum the known numbers: 25 + 35 + 28 + 40 = 128.
    3. Find the missing number: Let the fifth number be 'x'. The total sum is 128 + x = 150.
      x = 150 - 128 = 22. The fifth number is 22.

    6. For {55, 65, 70, 72, 75, 80, 82, 95, 100}, if the lowest score (55) is removed, which measure (mean or median) will change more?

    1. Original values:
      Original Mean = (55+65+70+72+75+80+82+95+100) / 9 = 694 / 9 ≈ 77.11
      Original Median (5th value of 9): 75
    2. New dataset after removing 55: {65, 70, 72, 75, 80, 82, 95, 100}. There are now 8 values.
    3. New values:
      New Mean = (694 - 55) / 8 = 639 / 8 = 79.875
      New Median (average of 4th and 5th values): (75 + 80) / 2 = 77.5
    4. Compare the changes:
      Change in Mean = 79.875 - 77.11 = 2.765
      Change in Median = 77.5 - 75 = 2.5
      The mean changed more than the median. This highlights that removing a value, especially one at the extreme end, has a greater impact on the mean. These foundational concepts are essential before moving on to topics like hypothesis testing.

    7. For house prices { 250 k , 250k, 275k, 310 k , 310k, 295k, 450 k , 450k, 310k, 1.2M}, which measure, mean or median, is more appropriate?</strong></p> <ol> <li><strong>Identify the outlier:</strong> The dataset contains a significant outlier:1,200,000. This price is much higher than the others.

  3. Calculate the mean: Sum = (250+275+310+295+450+310+1200)k = 3090 k . M e a n = 3090k. Mean = 3090k / 7 ≈ 441,428.</li> <li><strong>Calculate the median:</strong> First, order the prices: {250k, 275 k , 275k, 295k, 310 k , 310k, 310k, 450 k , 450k, 1.2M}. The median is the 4th value, which is 310 , 000. < / l i > < l i > < s t r o n g > C o n c l u s i o n : < / s t r o n g > T h e m e a n ( 310,000.</li> <li><strong>Conclusion:</strong> The mean ( 441,428) is heavily skewed by the 1.2 M h o u s e . T h e m e d i a n ( 1.2M house. The median ( 310,000) is a much better representation of the "typical" house price because it is not affected by the extreme outlier. For a deep dive into data properties, check out Khan Academy's unit on summarizing quantitative data.

    4. 8. The mean weight of 10 basketball players is 200 lbs. If the heaviest (240 lbs) and lightest (160 lbs) are removed, what is the new mean weight?

    1. Find the total original weight: Sum = Mean × Count = 200 lbs × 10 = 2000 lbs.
    2. Find the sum of the removed weights: 240 + 160 = 400 lbs.
    3. Calculate the new total weight: New Sum = 2000 - 400 = 1600 lbs.
    4. Calculate the new mean: After removing two players, there are 8 remaining. New Mean = 1600 lbs / 8 = 200 lbs. In this specific case, because the two removed values were equidistant from the original mean, the new mean is the same as the old one.

    9. You have {10, 15, 20, 25, 30}. If you add 100, how does the median change compared to the mean?

    1. Original values:
      Mean = (10+15+20+25+30) / 5 = 100 / 5 = 20
      Median (3rd value): 20
    2. New dataset with outlier: {10, 15, 20, 25, 30, 100}
    3. New values:
      New Mean = (100 + 100) / 6 = 200 / 6 ≈ 33.33
      New Median (average of 3rd and 4th values): (20 + 25) / 2 = 22.5
    4. Comparison: The mean increased from 20 to 33.33 (a change of 13.33). The median increased from 20 to 22.5 (a change of 2.5). The mean changes significantly more than the median when an outlier is introduced.

    10. The mode of {12, 18, 11, 25, 18, 12, x, 25} is 18. What is the smallest possible integer value for x?

    1. Count frequencies of existing numbers:
      11: 1 time
      12: 2 times
      18: 2 times
      25: 2 times
    2. Analyze the condition: The problem states that the mode is 18. This means 18 must be the most frequent number. Currently, 12, 18, and 25 are all tied with a frequency of 2.
    3. Determine the value of x: For 18 to be the unique mode, its frequency must be higher than any other number. Therefore, x must be 18, making its frequency 3.
    4. Answer: The value of x must be 18. The question asks for the smallest possible integer value, and since x needs to be 18 to make it the mode, the answer is 18.

    Quick Quiz

    Interactive Quiz 5 questions

    1. For the dataset {10, 20, 30, 40, 150}, which statement is true?

    • A The median is greater than the mean.
    • B The mean is greater than the median.
    • C The mean and median are equal.
    • D The mode is the most representative measure.
    Check answer

    Answer: B. The mean is greater than the median.

    2. How is the median of a dataset with 20 values calculated?

    • A It is the 10th value in the ordered list.
    • B It is the 11th value in the ordered list.
    • C It is the average of the 10th and 11th values in the ordered list.
    • D It is the average of the 9th and 10th values in the ordered list.
    Check answer

    Answer: C. It is the average of the 10th and 11th values in the ordered list.

    3. A dataset of student heights has two modes: 65 inches and 68 inches. What is this dataset called?

    • A Unimodal
    • B Bimodal
    • C Multimodal
    • D Non-modal
    Check answer

    Answer: B. Bimodal

    4. The mean of a dataset is 50. If every number in the dataset is increased by 5, what is the new mean?

    • A 50
    • B 55
    • C 250
    • D Cannot be determined
    Check answer

    Answer: B. 55

    5. Which measure of central tendency is always one of the actual data points in the set?

    • A Mean
    • B Median
    • C Mode
    • D All of the above
    Check answer

    Answer: C. Mode

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    Frequently Asked Questions

    When should I use the mean vs. the median?

    You should use the mean when your data is symmetrically distributed and does not have significant outliers. Use the median when your dataset is skewed or contains outliers, as the median provides a more robust and representative measure of the central value in these cases.

    Can a dataset have no mode?

    Yes, a dataset can have no mode. This occurs when all values in the dataset appear with the same frequency, typically just once. For example, in the set {1, 2, 3, 4, 5}, no value is more frequent than any other, so there is no mode.

    What is the relationship between mean, median, and mode in a skewed distribution?

    In a skewed distribution, the mean, median, and mode are not equal. For a positively skewed (right-skewed) distribution, the tail is on the right, and the relationship is typically Mean > Median > Mode. For a negatively skewed (left-skewed) distribution, the tail is on the left, and the relationship is Mean < Median < Mode. You can read more about this on the Wikipedia page for Skewness.

    How do I find the median of a large dataset?

    To find the median of a large dataset, first, you must sort all 'n' data points in ascending order. Then, use the formula (n + 1) / 2 to find the position of the median value. If 'n' is odd, this gives you the exact position of the median. If 'n' is even, this will result in a .5 value (e.g., 50.5), meaning you need to average the two data points at the positions on either side (e.g., the 50th and 51st values).

    Why are mean, median, and mode called measures of central tendency?

    They are called measures of central tendency because each one identifies a central, or typical, value that represents the middle of a dataset. They provide a single value summary that describes where the bulk of the data is located or clustered, offering a snapshot of the dataset's distribution.

    Does calculating mean, median, and mode relate to other statistical concepts?

    Absolutely. Mean, median, and mode are foundational concepts in descriptive statistics. They are often calculated alongside measures of dispersion or spread, such as variance and standard deviation, to provide a more complete picture of the data's characteristics. The mean is also a critical component in many advanced statistical tests and models.

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