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Statistics and Probability - Representation of data: frequency tables, histograms, and cumulative frequency graphs

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Frequency tables and the midpoint method: When data is grouped into classes like 10≤x<2010 \le x < 20, the midpoint x=10+202=15x = \frac{10+20}{2} = 15 is used as a representative value to estimate the mean using xˉ=∑f⋅x∑f\bar{x} = \frac{\sum f \cdot x}{\sum f}.

A simplified frequency table layout showing how the midpoint is derived from a class interval.
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Histograms use Frequency Density on the y-axis: The area of the bar represents the frequency. This is essential when class widths are unequal. Frequency Density=FrequencyClass Width\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}.

A histogram with unequal class widths showing varying bar heights based on frequency density.
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Cumulative Frequency Graphs: Data is plotted at the upper bound of each class. The graph is an 'S-curve' used to estimate the median (50%50\% mark) and the Interquartile Range (IQR).

Cumulative frequency curve showing how to find the median value on the x-axis from the 50% frequency mark.
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Interquartile Range (IQR): This represents the middle 50%50\% of the data. IQR=Q3−Q1IQR = Q_3 - Q_1, where Q1Q_1 is the lower quartile (25%25\% position) and Q3Q_3 is the upper quartile (75%75\% position).

📐Formulae

Estimated Mean (xˉ)=∑(f⋅x)∑f\text{Estimated Mean } (\bar{x}) = \frac{\sum (f \cdot x)}{\sum f}

Frequency Density=FrequencyClass Width\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}

Class Width=Upper Boundary−Lower Boundary\text{Class Width} = \text{Upper Boundary} - \text{Lower Boundary}

Interquartile Range (IQR)=Q3−Q1\text{Interquartile Range (IQR)} = Q_3 - Q_1

Lower Quartile Position (Q1)≈14n\text{Lower Quartile Position } (Q_1) \approx \frac{1}{4}n

Upper Quartile Position (Q3)≈34n\text{Upper Quartile Position } (Q_3) \approx \frac{3}{4}n

💡Examples

Problem 1:

Calculate the estimated mean for the following frequency table of test scores:

  • 0≤s<100 \le s < 10: Frequency 33
  • 10≤s<2010 \le s < 20: Frequency 88
  • 20≤s<3020 \le s < 30: Frequency 99

Solution:

  1. Find midpoints (xx) for each class:

    • Class 1: 0+102=5\frac{0+10}{2} = 5
    • Class 2: 10+202=15\frac{10+20}{2} = 15
    • Class 3: 20+302=25\frac{20+30}{2} = 25

  2. Calculate f⋅xf \cdot x for each class:

    • 3×5=153 \times 5 = 15
    • 8×15=1208 \times 15 = 120
    • 9×25=2259 \times 25 = 225

  3. Sum the frequencies (∑f\sum f): 3+8+9=203 + 8 + 9 = 20

  4. Sum the f⋅xf \cdot x values (∑f⋅x\sum f \cdot x): 15+120+225=36015 + 120 + 225 = 360

  5. Calculate Mean: xˉ=36020=18\bar{x} = \frac{360}{20} = 18

Explanation:

To estimate the mean from grouped data, we assume every value in an interval is equal to the midpoint of that interval. We then multiply these midpoints by their respective frequencies and divide by the total number of observations.

Problem 2:

A cumulative frequency graph for 8080 students' heights starts at (140,0)(140, 0) and ends at (190,80)(190, 80). If the curve passes through the point (170,60)(170, 60), what percentage of students are taller than 170170 cm?

Solution:

  1. Identify total number of students (nn): n=80n = 80.
  2. Identify students with height ≤170\le 170 cm: The yy-value at x=170x = 170 is 6060. This means 6060 students are 170170 cm or shorter.
  3. Calculate students taller than 170170 cm: 80−60=2080 - 60 = 20.
  4. Calculate as a percentage: 2080×100%=25%\frac{20}{80} \times 100\% = 25\%.

Explanation:

Cumulative frequency graphs always show the number of data points 'less than or equal to' a specific value. To find the number of values 'greater than', subtract the yy-value from the total frequency.

Problem 3:

A histogram is drawn for the distribution of the masses of 120120 parcels. One bar has a class interval of 10<m≤2510 < m \le 25 and a frequency of 4545. Calculate the frequency density for this bar.

A single histogram bar representing a frequency of 45 with a width of 15 and height of 3.

Solution:

Class Width=25−10=15\text{Class Width} = 25 - 10 = 15 Frequency Density=FrequencyClass Width\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} Frequency Density=4515=3\text{Frequency Density} = \frac{45}{15} = 3

Explanation:

To find the frequency density, we first determine the width of the class interval and then divide the frequency by that width. In a histogram, the height of the bar corresponds to this density.

Problem 4:

From the cumulative frequency curve provided, estimate the Interquartile Range (IQR) if Q1Q_1 occurs at x=22x = 22 and Q3Q_3 occurs at x=48x = 48.

Cumulative frequency curve indicating the positions of Q1 and Q3 on the horizontal axis.

Solution:

IQR=Q3−Q1IQR = Q_3 - Q_1 IQR=48−22IQR = 48 - 22 IQR=26IQR = 26

Explanation:

The IQR is the difference between the upper quartile and the lower quartile values on the x-axis (the data values, not the frequencies).