Solved Exercises on Histograms in Grade 7

Master histograms: frequency distribution, class intervals, frequency density, and data interpretation through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Histogram Construction
Exercise 1
The heights (in cm) of 20 students are: 145, 150, 155, 148, 152, 147, 153, 149, 151, 146, 154, 148, 150, 149, 152, 147, 151, 146, 150, 148. Construct a histogram using class intervals of 5 cm starting from 145 cm.
Definition:

Histogram: A graphical representation showing the distribution of data using bars where the area of each bar represents the frequency of data in each interval.

Class Interval: A range of values that groups data points together (e.g., 145-149 cm).

Histogram construction method:
  1. Organize data in ascending order
  2. Define class intervals of equal width
  3. Count the frequency of data in each interval
  4. Create a frequency table
  5. Draw the histogram with class intervals on x-axis and frequencies on y-axis
Step 1: Organize the data

Sorted data: 145, 146, 146, 147, 147, 148, 148, 148, 149, 149, 150, 150, 150, 151, 151, 152, 152, 153, 154, 155

Step 2: Define class intervals

Class intervals of 5 cm starting from 145 cm: 145-149, 150-154, 155-159

Step 3: Count frequencies

145-149: 8 students (145, 146, 146, 147, 147, 148, 148, 148, 149, 149)

150-154: 11 students (150, 150, 150, 151, 151, 152, 152, 153, 154)

155-159: 1 student (155)

Step 4: Create frequency table
Height Range (cm)Frequency
145-1498
150-15411
155-1591
Final answer:

The histogram has 3 bars: one for 145-149 cm with height 8, one for 150-154 cm with height 11, and one for 155-159 cm with height 1.

Applied rules:

• Equal intervals: Class intervals must be of equal width

• Frequency: Count of data points in each interval

• Area: In histograms, area represents frequency, not just height

145-149 150-154 155-159 0 5 10 15 8 11 1 Student Height Distribution
2 Frequency Density Histogram
Exercise 2
The test scores of 50 students are distributed as follows: 0-20 (frequency 5), 21-30 (frequency 10), 31-45 (frequency 15), 46-60 (frequency 20). Construct a histogram using frequency density since intervals are unequal.
Definition:

Frequency Density: Frequency per unit of class width = Frequency ÷ Class Width

Unequal Intervals: When class intervals have different widths, use frequency density to make fair comparisons.

Step 1: Calculate class widths

0-20: width = 21, 21-30: width = 10, 31-45: width = 15, 46-60: width = 15

Step 2: Calculate frequency densities

0-20: 5÷21 ≈ 0.24, 21-30: 10÷10 = 1.00, 31-45: 15÷15 = 1.00, 46-60: 20÷15 ≈ 1.33

Step 3: Create frequency density table
Score RangeFrequencyClass WidthFrequency Density
0-205210.24
21-3010101.00
31-4515151.00
46-6020151.33
Final answer:

The histogram uses frequency density as the y-axis to account for unequal class widths, ensuring accurate representation of data density across different intervals.

Applied rules:

• Frequency Density: Used when class intervals have unequal widths

• Area Proportionality: Area of each bar represents actual frequency

• Y-axis: Shows frequency density, not raw frequency

0-20 21-30 31-45 46-60 0 1 2 Test Scores Distribution (Frequency Density)
3 Data Interpretation
Exercise 3
A histogram shows the ages of people attending a concert. The age groups are: 10-19 (15 people), 20-29 (25 people), 30-39 (30 people), 40-49 (20 people), 50-59 (10 people). How many people attended? What is the modal class?
Definition:

Modal Class: The class interval with the highest frequency.

Total Frequency: Sum of all frequencies equals total number of observations.

Step 1: Identify the data

Age groups: 10-19 (15), 20-29 (25), 30-39 (30), 40-49 (20), 50-59 (10)

Step 2: Calculate total attendance

Total = 15 + 25 + 30 + 20 + 10 = 100 people

Step 3: Identify the modal class

The class with highest frequency is 30-39 with 30 people

Step 4: Analyze the distribution

The largest group is 30-39 years old (30%), followed by 20-29 (25%)

Final answer:

100 people attended the concert. The modal class is 30-39 years old.

Applied rules:

• Total calculation: Sum all frequencies to get total count

• Modal identification: Look for highest frequency bar

• Data interpretation: Understand what the histogram reveals about the dataset

10-19 20-29 30-39 40-49 50-59 0 20 40 15 25 30 20 10 Concert Attendee Age Distribution
Histogram Concepts, Rules and Methods
Frequency Density = \(\frac{Frequency}{Class Width}\)
Frequency Density Formula
Class Interval
Upper Limit - Lower Limit + 1
Range of values in each bar
Modal Class
Highest Frequency Interval
Most common data range
Total Frequency
Σ All Frequencies
Sum of all observations
Key definitions:

Histogram: A graphical representation of continuous data using bars where the area of each bar represents the frequency.

Class Interval: A range of values that groups data points together.

Frequency: The number of data points within each class interval.

Frequency Density: Frequency per unit of class width, used when intervals are unequal.

Histogram construction steps:
  1. Organize data: Sort values in ascending order
  2. Define intervals: Choose appropriate class intervals of equal width
  3. Count frequencies: Tally data points in each interval
  4. Create table: Record intervals and their frequencies
  5. Draw histogram: Plot intervals on x-axis, frequencies on y-axis
Tip 1: Always use equal class intervals unless specifically instructed otherwise.
Tip 2: When intervals are unequal, use frequency density to maintain accuracy.
Tip 3: Bars in histograms touch each other, unlike bar charts.
Tip 4: The modal class is the interval with the highest bar.
Key characteristics: Continuous data, equal intervals, area represents frequency, touching bars.
Common applications: Test scores, heights, weights, temperatures, time intervals.
Solution: Exercises 4 to 5
4 Finding Missing Values
Exercise 4
A histogram shows daily rainfall (mm) over 30 days. The intervals are: 0-5 (frequency 8), 6-10 (frequency ?), 11-15 (frequency 7), 16-20 (frequency 3). If the total is 30 days, find the missing frequency and draw the histogram.
Definition:

Missing Frequency: Calculated by subtracting known frequencies from total frequency.

Step 1: Set up equation

Known frequencies: 8 + ? + 7 + 3 = 30

18 + ? = 30

Step 2: Solve for missing frequency

? = 30 - 18 = 12

Step 3: Verify solution

8 + 12 + 7 + 3 = 30 ✓

Step 4: Create completed table
Rainfall (mm)Frequency
0-58
6-1012
11-157
16-203
Final answer:

The missing frequency is 12. The histogram has bars of heights 8, 12, 7, and 3 respectively.

Applied rules:

• Sum property: Total frequency equals sum of all individual frequencies

• Algebraic approach: Use equations to solve for unknowns

• Verification: Always check that calculated values satisfy the total

0-5 6-10 11-15 16-20 0 10 20 8 12 7 3 Daily Rainfall Distribution
5 Cumulative Analysis
Exercise 5
Using the histogram data from Exercise 1 (heights), determine how many students are taller than 149 cm and what percentage this represents of the total.
Definition:

Cumulative Frequency: Running total of frequencies up to a certain point.

Percentage Calculation: (Part ÷ Whole) × 100%

Step 1: Recall Exercise 1 data

145-149: 8 students, 150-154: 11 students, 155-159: 1 student

Step 2: Count students > 149 cm

Students taller than 149 cm: 11 + 1 = 12 students

Step 3: Calculate percentage

Percentage = (12 ÷ 20) × 100% = 60%

Step 4: Verify calculation

Students ≤ 149 cm: 8, Students > 149 cm: 12, Total: 8 + 12 = 20 ✓

Final answer:

12 students are taller than 149 cm, representing 60% of the total.

Applied rules:

• Cumulative counting: Add frequencies of relevant intervals

• Percentage formula: (part ÷ whole) × 100%

• Verification: Check that parts sum to whole

≤149 cm 150-154 cm 155-159 cm 0 5 10 15 8 11 1 Height Analysis: ≤149 vs >149 cm
Histogram Theory: Laws, Methods, Definitions, and Formulas
Frequency Density = \(\frac{Frequency}{Class Width}\)
Frequency Density
Key definitions:

Histogram: A graphical representation of continuous data using adjacent rectangles where the area of each rectangle is proportional to the frequency of the data in that interval.

Class Interval: A range of values that groups data points together (e.g., 0-10, 11-20).

Frequency: The number of data points falling within each class interval.

Modal Class: The class interval with the highest frequency.

Histogram creation methodology:
  1. Data Collection: Gather all data points to be represented
  2. Interval Selection: Choose appropriate class intervals of equal width
  3. Frequency Counting: Count data points in each interval
  4. Table Creation: Organize intervals and frequencies in a table
  5. Graph Construction: Draw adjacent rectangles with heights proportional to frequencies
Tip 1: Always ensure class intervals are mutually exclusive and collectively exhaustive.
Tip 2: Bars in histograms must touch each other to represent continuous data.
Tip 3: For unequal intervals, use frequency density to maintain accurate area representation.
Tip 4: The total area of all bars equals the total frequency of the dataset.
Key characteristics: Continuous data representation, adjacent bars, area represents frequency, suitable for large datasets.
Common applications: Statistical analysis, quality control, scientific research, demographic studies.
Essential formulas and rules:

• Class Width: Upper limit - Lower limit + 1

• Frequency Density: Frequency ÷ Class Width (when intervals are unequal)

• Total Frequency: Sum of all individual frequencies

• Percentage: (Individual frequency ÷ Total frequency) × 100%

• Modal Class: Class with maximum frequency

Questions & Answers

Question: I don't understand why we sometimes use frequency density instead of just frequency in histograms. When do I know which one to use?

Answer: Great question! The choice depends on the class intervals:

  • When all class intervals have the same width, you can use regular frequency on the y-axis.
  • When class intervals have different widths, you must use frequency density to ensure accurate representation.

Here's why: If you have intervals like 0-10 (width 11) and 11-15 (width 5), and both have frequency 10, the wider interval would appear to have more data if we only looked at frequency. Frequency density (frequency ÷ class width) normalizes this difference.

Example: Interval 0-10 with frequency 20 has density 20÷11 ≈ 1.82, while interval 11-15 with frequency 10 has density 10÷5 = 2.0. This shows that the narrower interval actually has higher density.

Question: How do I decide what class intervals to use when I'm given raw data for a histogram?

Answer: Here's a systematic approach to choosing class intervals:

  • Range calculation: Find the difference between the highest and lowest values
  • Number of intervals: Aim for 5-10 intervals depending on dataset size
  • Interval width: Round the range ÷ number of intervals to a convenient number
  • Start point: Choose a round number slightly below the minimum value

Example: If data ranges from 12 to 98, range = 86. For 8 intervals, width ≈ 11. Round to 10. Start at 10, giving intervals: 10-19, 20-29, ..., 90-99.

Key principle: Choose intervals that are easy to work with (multiples of 5 or 10) while capturing the data distribution effectively.

Question: What's the difference between a histogram and a bar chart? They look similar to me.

Answer: While they may look similar, histograms and bar charts serve different purposes:

  • Histogram: Represents continuous data; bars touch each other; area represents frequency; used for distribution analysis
  • Bar Chart: Represents categorical data; bars have gaps between them; height represents frequency; used for comparison

In histograms, the x-axis shows continuous numerical ranges (like 0-10, 11-20). In bar charts, the x-axis shows distinct categories (like apple, orange, banana).

Think of it this way: histograms show how data is distributed across ranges, while bar charts compare quantities across different categories.

Question: How do I find the median from a histogram? It seems impossible since I don't have exact values.

Answer: You can estimate the median from a histogram using cumulative frequency:

  1. Calculate total frequency (n) and find n/2
  2. Create a cumulative frequency table
  3. Identify which class interval contains the n/2th value
  4. Use interpolation within that interval to estimate the median

For example, if total frequency is 50, the median is the 25th value. Find which interval contains the 25th cumulative frequency, then interpolate within that interval based on its width and frequency.

This gives an approximation since we don't know the exact distribution within each interval.

Question: What does it mean when a histogram is skewed to the left or right? How can I tell?

Answer: Skewness describes the asymmetry of a distribution:

  • Right-skewed (positive skew): Tail extends to the right; mean > median; occurs when there are unusually high values
  • Left-skewed (negative skew): Tail extends to the left; mean < median; occurs when there are unusually low values
  • Symmetric: Both sides are mirror images; mean ≈ median

To identify skewness, look at the longer tail of the distribution. If the histogram has a longer tail on the right side, it's right-skewed. If the longer tail is on the left, it's left-skewed.

Example: Income distribution is typically right-skewed because most people earn moderate incomes, but a few earn extremely high incomes, creating a long right tail.