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Statistics and Probability - Frequency tables – Grouped Data

Grade 12IB_AA

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

🔑Concepts

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Grouped data is used when the data set contains a large number of values or when the data is continuous. Data is organized into class intervals (e.g., 10≤x<2010 \le x < 20).

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The midpoint (xix_i) of a class interval is calculated as the average of the upper and lower boundaries: xi=lower boundary+upper boundary2x_i = \frac{\text{lower boundary} + \text{upper boundary}}{2}.

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For grouped data, we assume all values in an interval are represented by the midpoint. This results in an estimated mean rather than an exact mean.

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The modal class is the class interval with the highest frequency (ff).

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The median class is the interval that contains the middle value, often found by looking at the cumulative frequency reaching n2\frac{n}{2}.

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Cumulative frequency is the running total of frequencies, used to construct a cumulative frequency curve (ogive) to estimate the median and quartiles.

📐Formulae

Estimated Mean xˉ=∑i=1kfixin\text{Estimated Mean } \bar{x} = \frac{\sum_{i=1}^{k} f_i x_i}{n}

Total Frequency n=∑i=1kfi\text{Total Frequency } n = \sum_{i=1}^{k} f_i

Estimated Variance σ2=∑fi(xi−xˉ)2n=∑fixi2n−xˉ2\text{Estimated Variance } \sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{n} = \frac{\sum f_i x_i^2}{n} - \bar{x}^2

Estimated Standard Deviation σ=∑fixi2n−xˉ2\text{Estimated Standard Deviation } \sigma = \sqrt{\frac{\sum f_i x_i^2}{n} - \bar{x}^2}

💡Examples

Problem 1:

The table below shows the heights (in cm) of 40 plants. Calculate an estimate for the mean height.

Height (h)Frequency (f)0≤h<10510≤h<201220≤h<301830≤h<405\begin{array}{|c|c|} \hline \text{Height } (h) & \text{Frequency } (f) \\ \hline 0 \le h < 10 & 5 \\ 10 \le h < 20 & 12 \\ 20 \le h < 30 & 18 \\ 30 \le h < 40 & 5 \\ \hline \end{array}

Solution:

  1. Find the midpoints (xix_i): 5,15,25,355, 15, 25, 35.
  2. Calculate fixif_i x_i for each row:
  • 5×5=255 \times 5 = 25
  • 12×15=18012 \times 15 = 180
  • 18×25=45018 \times 25 = 450
  • 5×35=1755 \times 35 = 175
  1. Sum the frequencies: n=5+12+18+5=40n = 5 + 12 + 18 + 5 = 40.
  2. Sum the products: ∑fixi=25+180+450+175=830\sum f_i x_i = 25 + 180 + 450 + 175 = 830.
  3. Calculate the mean: xˉ=83040=20.75 cm\bar{x} = \frac{830}{40} = 20.75 \text{ cm}

Explanation:

To estimate the mean from a grouped frequency table, find the midpoint of each interval, multiply by the frequency, sum those products, and divide by the total number of observations (nn).

Problem 2:

Identify the modal class and the median class for the data in the previous example.

Solution:

  1. Modal Class: The highest frequency is 1818, which corresponds to the interval 20≤h<3020 \le h < 30.
  2. Median Class: The total frequency is n=40n = 40. The median is the 402=20th\frac{40}{2} = 20^{th} value.
  • Cumulative frequency for 0≤h<100 \le h < 10: 55
  • Cumulative frequency for 10≤h<2010 \le h < 20: 5+12=175 + 12 = 17
  • Cumulative frequency for 20≤h<3020 \le h < 30: 17+18=3517 + 18 = 35 Since the 20th20^{th} value falls within the 20≤h<3020 \le h < 30 interval, that is the median class.

Explanation:

The modal class is the interval with the highest frequency. The median class is the interval where the cumulative frequency first reaches or exceeds half of the total population (n/2n/2).