Mean Median And Mode Practice Problems

Let's dive into the world of mean, median, and mode with a series of practice problems designed to sharpen your understanding and boost your confidence. These statistical measures are fundamental tools in data analysis, helping us to summarize and interpret information effectively.

Understanding Mean, Median, and Mode

Before tackling the practice problems, it’s crucial to have a solid grasp of what each term represents:

• Mean: The average of a dataset. Calculated by summing all the values and dividing by the total number of values.
• Median: The middle value in a dataset when it's ordered from least to greatest. If there's an even number of values, the median is the average of the two middle values.
• Mode: The value that appears most frequently in a dataset. A dataset can have one mode (unimodal), multiple modes (bimodal, trimodal, etc.), or no mode if all values appear with the same frequency.

Practice Problems: Level 1 (Basic)

Let's start with some basic examples to solidify your understanding of each concept.

Problem 1:

Find the mean, median, and mode of the following dataset: 4, 7, 2, 9, 5, 7, 3.

Solution:

1. Mean:
   • Sum of values: 4 + 7 + 2 + 9 + 5 + 7 + 3 = 37
   • Number of values: 7
   • Mean = 37 / 7 = 5.29 (approximately)
2. Median:
   • First, order the dataset: 2, 3, 4, 5, 7, 7, 9
   • The middle value is 5.
   • Median = 5
3. Mode:
   • The value 7 appears twice, which is more frequent than any other value.
   • Mode = 7

Problem 2:

Calculate the mean, median, and mode for this dataset: 12, 15, 11, 18, 12, 14, 17.

Solution:

1. Mean:
   • Sum of values: 12 + 15 + 11 + 18 + 12 + 14 + 17 = 99
   • Number of values: 7
   • Mean = 99 / 7 = 14.14 (approximately)
2. Median:
   • Order the dataset: 11, 12, 12, 14, 15, 17, 18
   • The middle value is 14.
   • Median = 14
3. Mode:
   • The value 12 appears twice, which is more frequent than any other value.
   • Mode = 12

Problem 3:

Determine the mean, median, and mode of the data: 25, 28, 30, 25, 27, 32.

Solution:

1. Mean:
   • Sum of values: 25 + 28 + 30 + 25 + 27 + 32 = 167
   • Number of values: 6
   • Mean = 167 / 6 = 27.83 (approximately)
2. Median:
   • Order the dataset: 25, 25, 27, 28, 30, 32
   • Since there's an even number of values, the median is the average of the two middle values (27 and 28).
   • Median = (27 + 28) / 2 = 27.5
3. Mode:
   • The value 25 appears twice, which is more frequent than any other value.
   • Mode = 25

Practice Problems: Level 2 (Intermediate)

Now, let's increase the complexity with larger datasets and scenarios.

Problem 4:

The following are the test scores of 10 students: 75, 80, 85, 90, 95, 80, 70, 85, 90, 100. Find the mean, median, and mode of these scores.

Solution:

1. Mean:
   • Sum of values: 75 + 80 + 85 + 90 + 95 + 80 + 70 + 85 + 90 + 100 = 850
   • Number of values: 10
   • Mean = 850 / 10 = 85
2. Median:
   • Order the dataset: 70, 75, 80, 80, 85, 85, 90, 90, 95, 100
   • Since there's an even number of values, the median is the average of the two middle values (85 and 85).
   • Median = (85 + 85) / 2 = 85
3. Mode:
   • The values 80, 85, and 90 each appear twice.
   • Mode = 80, 85, 90 (This dataset is trimodal)

Problem 5:

A company has 15 employees. Their salaries are as follows (in thousands of dollars): 40, 45, 50, 55, 60, 40, 45, 50, 55, 65, 40, 50, 60, 70, 80. Calculate the mean, median, and mode of the salaries.

Solution:

1. Mean:
   • Sum of values: 40 + 45 + 50 + 55 + 60 + 40 + 45 + 50 + 55 + 65 + 40 + 50 + 60 + 70 + 80 = 805
   • Number of values: 15
   • Mean = 805 / 15 = 53.67 (approximately)
2. Median:
   • Order the dataset: 40, 40, 40, 45, 45, 50, 50, 50, 55, 55, 60, 60, 65, 70, 80
   • The middle value is 50.
   • Median = 50
3. Mode:
   • The value 40 appears three times, and the value 50 appears three times.
   • Mode = 40, 50 (This dataset is bimodal)

Problem 6:

The number of customers visiting a store each day for a week is: 120, 150, 130, 140, 120, 160, 150. Find the mean, median, and mode of the daily customer count.

Solution:

1. Mean:
   • Sum of values: 120 + 150 + 130 + 140 + 120 + 160 + 150 = 970
   • Number of values: 7
   • Mean = 970 / 7 = 138.57 (approximately)
2. Median:
   • Order the dataset: 120, 120, 130, 140, 150, 150, 160
   • The middle value is 140.
   • Median = 140
3. Mode:
   • The value 120 appears twice, and the value 150 appears twice.
   • Mode = 120, 150 (This dataset is bimodal)

Practice Problems: Level 3 (Advanced)

These problems require a deeper understanding and might involve some algebraic thinking.

Problem 7:

The mean of 5 numbers is 20. If four of the numbers are 15, 18, 22, and 25, what is the fifth number?

Solution:

• Let the fifth number be x.
• The sum of the 5 numbers is 5 * 20 = 100.
• 15 + 18 + 22 + 25 + x = 100
• 80 + x = 100
• x = 100 - 80
• x = 20

Therefore, the fifth number is 20.

Problem 8:

A dataset consists of 6 numbers. The median is 15, and the numbers are 10, 12, x, y, 18, and 20, where x < y. Find the possible values of x and y.

Solution:

• Order the dataset: 10, 12, x, y, 18, 20
• Since there are 6 numbers, the median is the average of the two middle values, x and y.
• ( x + y ) / 2 = 15
• x + y = 30
• Since x < y, we need to find two numbers that add up to 30, where x is greater than 12 (because of the order) and y is less than 18.

Possible values:

• x = 13, y = 17
• x = 14, y = 16
• x = 15, y = 15 (While technically valid, the problem states x < y, so this isn't ideal, but can be considered a solution if the constraint isn't strictly enforced).

Problem 9:

The mode of the following dataset is 8: 5, 6, 8, 7, 8, 9, x. What is the minimum value of x that makes 8 the mode? What other values of x would also make 8 the mode?

Solution:

• Currently, 8 appears twice in the dataset. To make 8 the mode, x must be equal to 8.
• Therefore, the minimum value of x is 8.
• If x = 8, the dataset becomes 5, 6, 8, 7, 8, 9, 8, and 8 appears three times, making it the mode.

Problem 10:

A set of data has a mean of 10. If each data point is increased by 2, what is the new mean?

Solution:

Let the original data points be $x_1, x_2, ..., x_n$. The original mean is:

$\frac{x_1 + x_2 + ... + x_n}{n} = 10$

Now, each data point is increased by 2. The new data points are $x_1 + 2, x_2 + 2, ..., x_n + 2$. The new mean is:

$\frac{(x_1 + 2) + (x_2 + 2) + ... + (x_n + 2)}{n}$

We can rewrite this as:

$\frac{x_1 + x_2 + ... + x_n + 2n}{n}$

Which can be further separated into:

$\frac{x_1 + x_2 + ... + x_n}{n} + \frac{2n}{n}$

Since $\frac{x_1 + x_2 + ... + x_n}{n} = 10$, the new mean becomes:

$10 + 2 = 12$

Therefore, the new mean is 12.

Practice Problems: Level 4 (Challenging)

These problems require applying the concepts in more complex ways and might involve combining mean, median, and mode.

Problem 11:

The average of five consecutive integers is 21. What is the median of these integers?

Solution:

Let the five consecutive integers be n, n+1, n+2, n+3, and n+4.

The average of these integers is:

( n + (n+1) + (n+2) + (n+3) + (n+4) ) / 5 = 21

(5n + 10) / 5 = 21

5n + 10 = 105

5n = 95

n = 19

The five consecutive integers are 19, 20, 21, 22, and 23.

The median is the middle value, which is 21. Therefore, the median is 21. (In fact, for consecutive integers, the mean and median are always the same.)

Problem 12:

A list of numbers consists of the following: 2, 4, 6, 8, 10. A new number, x, is added to the list. If the new mean is equal to the new median, what is the value of x? Consider two cases: x < 6 and x > 6.

Solution:

First, find the original mean and median:

Original list: 2, 4, 6, 8, 10

Original mean: (2+4+6+8+10)/5 = 30/5 = 6

Original median: 6

Now consider the two cases:

• Case 1: x < 6

  The new ordered list will be x, 2, 4, 6, 8, 10. The new median will be (4+6)/2 = 5.

  The new mean is (2+4+6+8+10+ x)/6 = (30+ x)/6

  We are given that the new mean equals the new median: (30+ x)/6 = 5

  30 + x = 30

  x = 0

  Since 0 < 6, this is a valid solution.

• Case 2: x > 6

  The new ordered list will be 2, 4, 6, 8, 10, x. The new median will be (6+8)/2 = 7.

  The new mean is still (2+4+6+8+10+ x)/6 = (30+ x)/6

  We are given that the new mean equals the new median: (30+ x)/6 = 7

  30 + x = 42

  x = 12

  Since 12 > 6, this is a valid solution.

Therefore, the possible values of x are 0 and 12.

Problem 13:

A teacher gives a quiz to two classes. The first class has 20 students and their average score is 80. The second class has 30 students and their average score is 70. What is the overall average score of all students?

Solution:

• The total score of the first class is 20 * 80 = 1600
• The total score of the second class is 30 * 70 = 2100
• The total score of all students is 1600 + 2100 = 3700
• The total number of students is 20 + 30 = 50
• The overall average score is 3700 / 50 = 74

Therefore, the overall average score of all students is 74.

Problem 14:

The ages of 10 people are: 10, 12, 15, 18, 20, 22, 25, 28, 30, 35. Calculate the mean and median. If the oldest person (age 35) leaves the group, how do the mean and median change?

Solution:

• Original Data: 10, 12, 15, 18, 20, 22, 25, 28, 30, 35

• Original Mean: (10 + 12 + 15 + 18 + 20 + 22 + 25 + 28 + 30 + 35) / 10 = 215 / 10 = 21.5

• Original Median: (20 + 22) / 2 = 21

• Data After Removing 35: 10, 12, 15, 18, 20, 22, 25, 28, 30

• New Mean: (10 + 12 + 15 + 18 + 20 + 22 + 25 + 28 + 30) / 9 = 180 / 9 = 20

• New Median: 20

The mean decreases from 21.5 to 20, and the median decreases from 21 to 20. Removing the largest value has a noticeable impact on the mean and, in this case, a small impact on the median.

Problem 15:

The numbers 3, 5, 7, 9, and x have a mean of 6. Find the mode of the numbers 3, 5, 7, 9, x, 3, 5, 7, 9, x.

Solution:

• Step 1: Find the value of x. The mean of 3, 5, 7, 9, and x is 6. So we have: (3 + 5 + 7 + 9 + x) / 5 = 6 (24 + x) / 5 = 6 24 + x = 30 x = 6
• Step 2: Form the new set of numbers. Now we have the set of numbers: 3, 5, 7, 9, 6, 3, 5, 7, 9, 6
• Step 3: Find the mode of the new set. Let's count the frequency of each number: 3 appears 2 times 5 appears 2 times 6 appears 2 times 7 appears 2 times 9 appears 2 times Since each number appears twice, there is no unique mode. The dataset is multimodal (specifically, it's 5-modal). Therefore, the modes are 3, 5, 6, 7, and 9.

Conclusion

Working through these practice problems should significantly improve your understanding of mean, median, and mode. Remember to always carefully consider the context of the data and choose the most appropriate measure of central tendency to represent the information accurately. As you continue to practice, you'll become more comfortable and confident in your ability to analyze and interpret data effectively. Good luck!