Solved Exercises on Measures of Center in Grade 7

Master measures of center: mean, median, mode, and range through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Mean Calculation
Exercise 1
Find the mean of the following test scores: 85, 92, 78, 96, 88, 91, 83.
Definition:

Mean (Average): The sum of all values divided by the number of values.

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

Mean Calculation Method:
  1. Add all the values together
  2. Count the number of values
  3. Divide the sum by the number of values
  4. Round to appropriate decimal places if necessary
Sum
613
Count
7
Mean
87.6
Step 1: Add all values

85 + 92 + 78 + 96 + 88 + 91 + 83 = 613

Step 2: Count number of values

There are 7 test scores

Step 3: Divide sum by count

Mean = 613 ÷ 7 = 87.571428...

Step 4: Round appropriately

Mean ≈ 87.6 (rounded to one decimal place)

Mean = 87.6
Final Answer:

The mean of the test scores is 87.6.

Applied Rules:

• Mean Formula: Sum of values divided by count

• Arithmetic: Perform addition before division

• Rounding: Round to one more decimal than original data

2 Median Calculation
Exercise 2
Find the median of the following data set: 12, 18, 15, 21, 10, 16, 19, 14.
Definition:

Median: The middle value when data is arranged in ascending order.

For odd number of values: The middle value

For even number of values: Average of the two middle values

Sorted Data
10, 12, 14, 15, 16, 18, 19, 21
Middle Values
15, 16
Median
15.5
Step 1: Arrange data in ascending order

Original: 12, 18, 15, 21, 10, 16, 19, 14

Sorted: 10, 12, 14, 15, 16, 18, 19, 21

Step 2: Count the number of values

There are 8 values (even number)

Step 3: Identify the two middle values

Position 4: 15, Position 5: 16

These are the 4th and 5th values in the sorted list

Step 4: Calculate the median

Median = (15 + 16) ÷ 2 = 31 ÷ 2 = 15.5

Median = 15.5
Final Answer:

The median of the data set is 15.5.

Applied Rules:

• Sorting Required: Always arrange data in order first

• Odd Count: Median is the middle value

• Even Count: Median is average of two middle values

3 Mode Calculation
Exercise 3
Find the mode(s) of the following data set: 5, 7, 3, 8, 5, 2, 7, 5, 9, 7.
Definition:

Mode: The value that appears most frequently in a data set.

Unimodal: One mode exists

Bimodal: Two modes exist

Trimodal: Three modes exist

No Mode: All values appear equally often

Value Counts
5:3, 7:3, others:1
Highest Frequency
3 occurrences
Modes
5 and 7
Step 1: List all values

Data: 5, 7, 3, 8, 5, 2, 7, 5, 9, 7

Step 2: Count frequency of each value

2: appears 1 time

3: appears 1 time

5: appears 3 times

7: appears 3 times

8: appears 1 time

9: appears 1 time

Step 3: Identify the highest frequency

Values 5 and 7 both appear 3 times (highest frequency)

Step 4: Determine the mode(s)

Since two values tie for highest frequency, both are modes

Mode = 5 and 7 (bimodal)

Modes = 5 and 7 (bimodal)
Final Answer:

The modes of the data set are 5 and 7. This is a bimodal distribution.

Applied Rules:

• Frequency Count: Tally how many times each value appears

• Multiple Modes: More than one value can be a mode

• No Mode: If all values appear equally, no mode exists

Rules and methods, laws,...
\(\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}\)
Mean Formula
Mean (Average)
\(\bar{x} = \frac{\sum x}{n}\)
Sum divided by count
Median
Middle value when sorted
Depends on data count
Mode
Most frequent value
May have multiple or none
Key Definitions:

Mean: The arithmetic average of all values in a data set

Median: The middle value when data is arranged in ascending order

Mode: The value that appears most frequently in a data set

Range: The difference between the maximum and minimum values

Unimodal: A data set with one mode

Bimodal: A data set with two modes

Outlier: A value that is significantly different from other values

Complete Methodology:
  1. Mean Calculation: Add all values, divide by count
  2. Median Calculation: Sort data, find middle value(s)
  3. Mode Identification: Count frequencies, identify most common
  4. Range Calculation: Subtract minimum from maximum
  5. Outlier Detection: Look for extreme values that may affect measures
  6. Comparison: Compare measures to understand data distribution
Tip 1: Always sort data before finding the median.
Tip 2: Mean is affected by outliers, median is not.
Tip 3: A data set can have one, multiple, or no modes.
Tip 4: Mean uses all data points, median uses position, mode uses frequency.
Tip 5: When mean > median, data is skewed right; when mean < median, data is skewed left.
Common Errors: Forgetting to sort for median, miscounting frequencies, calculation mistakes, not considering outliers.
Exam Preparation: Practice with various data sets, memorize formulas, understand when each measure is most appropriate.
Solution: Exercises 4 to 5
4 Outliers and Their Effects
Exercise 4
Consider the data set: 15, 18, 20, 17, 16, 19, 18, 100. Calculate the mean and median with and without the outlier. Explain the effect.
Definition:

Outlier: A value that is significantly different from other values in the data set.

Effect on Mean: Outliers strongly affect the mean because it uses all data values.

Effect on Median: Outliers have less effect on the median since it depends on position.

With Outlier
Mean=27.9, Med=17.5
Without Outlier
Mean=17.7, Med=17.5
Effect
Mean increases by 10.2
Step 1: Identify the outlier

Value 100 is significantly larger than other values (15-20 range)

This is the outlier.

Step 2: Calculate measures with outlier

Mean with outlier: (15+18+20+17+16+19+18+100)÷8 = 203÷8 = 25.375 ≈ 25.4

Sorted data: 15, 16, 17, 18, 18, 19, 20, 100

Median: (18+18)÷2 = 18

Step 3: Calculate measures without outlier

Mean without outlier: (15+18+20+17+16+19+18)÷7 = 123÷7 = 17.57 ≈ 17.6

Sorted data: 15, 16, 17, 18, 18, 19, 20

Median: 18 (middle value)

Step 4: Compare the effects

Mean changed from 17.6 to 25.4 (increase of 7.8)

Median remained at 18 (no change)

Step 5: Explain the effect

The outlier significantly increased the mean but did not affect the median.

This shows that mean is sensitive to extreme values.

With outlier: Mean=25.4, Median=18
Without outlier: Mean=17.6, Median=18
Final Answer:

With outlier: Mean = 25.4, Median = 18. Without outlier: Mean = 17.6, Median = 18. The outlier significantly increased the mean but had no effect on the median.

Applied Rules:

• Outlier Sensitivity: Mean is sensitive to outliers, median is resistant

• Data Analysis: Always consider outliers when interpreting measures

• Robust Measures: Median is more robust than mean for skewed data

5 Choosing the Best Measure
Exercise 5
For each situation, explain which measure of center (mean, median, or mode) would be most appropriate and why.
a) Salaries at a company with one CEO earning much more than employees
b) Favorite ice cream flavor among students
c) Typical home prices in a neighborhood
Definition:

Appropriate Measure: The measure that best represents the typical value in a data set.

Skewed Data: Data that is not symmetric, with a tail extending in one direction.

Categorical Data: Data that represents categories rather than numerical values.

Situation a
Median
Situation b
Mode
Situation c
Median
Step 1: Analyze Situation a - Salaries with CEO

The CEO's salary is an extreme outlier that would skew the mean upward.

The median represents the typical employee salary better.

Step 2: Analyze Situation b - Ice Cream Flavors

This is categorical data, not numerical.

Only the mode makes sense (most popular flavor).

Step 3: Analyze Situation c - Home Prices

Home prices often have extreme outliers (luxury homes).

The median better represents typical home prices.

Step 4: Justify each choice

a) Median: Resistant to outliers, shows typical employee salary

b) Mode: Only appropriate for categorical data

c) Median: Less affected by luxury home outliers

Step 5: Consider data characteristics

Always consider skewness, data type, and presence of outliers when choosing.

a) Median, b) Mode, c) Median
Final Answer:

a) Median - resistant to CEO's high salary outlier
b) Mode - appropriate for categorical data
c) Median - resistant to luxury home outliers

Applied Rules:

• Data Type: Use mode for categorical data

• Outlier Presence: Use median when outliers exist

• Distribution Shape: Mean for symmetric, median for skewed

Detailed Summary: Measures of Center Fundamentals
\(\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}\)
Mean Formula
Key definitions:

Mean: The arithmetic average of all values in a data set, calculated by adding all values and dividing by the count

Median: The middle value when data is arranged in ascending order; if even number of values, it's the average of the two middle values

Mode: The value that appears most frequently in a data set; there can be one, multiple, or no modes

Range: The difference between the maximum and minimum values in a data set

Outlier: A data point that is significantly different from other values in the data set

Skewed Distribution: A distribution where data is not symmetric, with a longer tail on one side

Complete methodology:
  1. Mean Calculation: Add all values and divide by the number of values
  2. Median Calculation: Sort data in ascending order, then find the middle value(s)
  3. Mode Identification: Count the frequency of each value and identify the most frequent
  4. Range Calculation: Subtract the minimum value from the maximum value
  5. Outlier Analysis: Identify extreme values and assess their impact on measures
  6. Measure Selection: Choose the most appropriate measure based on data characteristics
Tip 1: Always sort data before finding the median to ensure accuracy.
Tip 2: The mean is sensitive to outliers, while the median is resistant to them.
Tip 3: For symmetric distributions, mean and median are approximately equal.
Tip 4: Use mode for categorical data or when looking for the most common value.
Tip 5: When mean > median, data is skewed right; when mean < median, data is skewed left.
Common errors: Forgetting to sort for median, miscounting frequencies, calculation mistakes, not considering the effect of outliers, choosing inappropriate measures for data type.
Exam preparation: Practice calculations with various data sets, memorize formulas, understand when each measure is most appropriate, know the effect of outliers on different measures.
Formulas to know by heart:

• Mean Formula: Mean = (Sum of all values) ÷ (Number of values)

• Median Rule: Middle value when sorted (average of two middle values if even count)

• Mode Rule: Value that appears most frequently

• Range Formula: Range = Maximum - Minimum

• Outlier Impact: Mean is sensitive, median is resistant

Exercise with Visualization: Comparing Measures
Exercise 6: Data Distribution Comparison
Consider the following data sets and their measures of center:
Set A: 10, 12, 14, 15, 16, 18, 20 (symmetric)
Set B: 5, 7, 9, 11, 13, 15, 50 (skewed right)
Set C: 2, 2, 3, 4, 5, 5, 5 (with mode)

Analysis: The visualization shows how different distributions affect measures of center.

  • Set A: Mean ≈ Median (symmetric distribution)
  • Set B: Mean > Median (skewed right by outlier)
  • Set C: Mode is prominent at value 5

Questions & Answers

Question: How do I remember which measure of center to use when? They all seem important.

Answer: Here's a simple decision tree to help you choose:

  • Is the data categorical? → Use MODE (e.g., favorite colors, types of pets)
  • Are there outliers or is the data skewed? → Use MEDIAN (e.g., salaries, home prices)
  • Is the data symmetric with no outliers? → Use MEAN (e.g., test scores, heights)

Think of it this way:

  • Mean: Uses all data points, good for symmetric distributions
  • Median: Represents the middle, unaffected by extreme values
    • Mode: Shows the most common value, works for any data type

Practice with different scenarios to build your intuition about which measure is most appropriate!

Question: Why does the median require sorting the data? Can't I just pick the middle number?

Answer: Sorting is essential for finding the median because:

  • Definition Requirement: The median is specifically the middle value when data is arranged in order
  • Position Matters: We need to know the exact position of the middle value
  • Accuracy: Without sorting, you might pick a value that isn't actually the middle

Example: In the set [15, 12, 18, 10, 16], if you don't sort, you might think 18 is the middle (position 3), but actually it's 15. When sorted [10, 12, 15, 16, 18], the middle value is clearly 15.

Sorting ensures you're finding the true middle value based on the data's magnitude, not its original position in the list.

Always sort first, then identify the middle position!

Question: Can a data set have more than one mode? What happens if every number appears once?

Answer: Yes, a data set can have multiple modes:

  • Unimodal: One mode (e.g., [1, 2, 2, 3] → mode is 2)
  • Bimodal: Two modes (e.g., [1, 1, 2, 3, 3] → modes are 1 and 3)
  • Trimodal: Three modes (e.g., [1, 1, 2, 2, 3, 3, 4] → modes are 1, 2, and 3)
  • No Mode: All values appear equally often (e.g., [1, 2, 3, 4, 5] → no mode)

If every number appears exactly once, there is no mode because no value appears more frequently than others.

It's important to note that unlike mean and median, a data set can have multiple modes or no mode at all. This makes mode unique among the measures of center.

When describing the mode, always specify if there are multiple modes or if there is no mode.

Question: How do outliers affect the mean and median differently? Why is this important?

Answer: The difference lies in how each measure is calculated:

  • Mean: Uses ALL values in its calculation (sum ÷ count), so outliers significantly affect it
  • Median: Depends only on the MIDDLE position when sorted, so outliers have minimal effect

Example: For [10, 12, 14, 16, 18], mean = median = 14

Adding outlier 100: [10, 12, 14, 16, 18, 100]

  • New mean = 170 ÷ 6 = 28.3 (increased significantly)
  • New median = (14+16) ÷ 2 = 15 (changed minimally)

This is important because it helps you choose the appropriate measure. When data has outliers, median often better represents the "typical" value. This is why median income is often reported instead of mean income!

Question: How can I check if I calculated the measures of center correctly?

Answer: Here are several verification strategies:

  1. Mean Check: Multiply your calculated mean by the count; it should equal the sum of original values
  2. Median Check: Ensure your sorted data has the correct number of values before and after the median
  3. Mode Check: Re-count the frequency of the mode you identified
  4. Reasonableness Check: Does your answer make sense in the context?

For mean: If you calculated mean = 85 for [80, 85, 90], verify: 85 × 3 = 255, and 80+85+90 = 255 ✓

For median: In [5, 10, 15, 20, 25], median should be 15, and there should be 2 values below and 2 above.

Always double-check your arithmetic and ensure you followed the correct procedure for each measure!