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Statistics and Probability - Histograms for continuous fixed interval groups-extended

Grade 10IB

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

🔑Concepts

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Histograms are used to represent continuous data where the data is grouped into intervals (classes). Unlike bar charts, there are no gaps between the bars because the data is continuous.

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For continuous data with fixed (equal) intervals, the height of each bar represents the frequency of that class interval.

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Class intervals are usually written in the form a≤x<ba \le x < b, where aa is the lower boundary and bb is the upper boundary.

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The class width is calculated as: Upper Boundary−Lower Boundary\text{Upper Boundary} - \text{Lower Boundary}.

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To estimate the mean from a grouped frequency table or histogram, we assume all values in an interval are at the midpoint of that interval.

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The area of each bar in a histogram is proportional to the frequency. When intervals are equal, the area is directly proportional to the height.

📐Formulae

Midpoint=Lower Boundary+Upper Boundary2\text{Midpoint} = \frac{\text{Lower Boundary} + \text{Upper Boundary}}{2}

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

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

Frequency=Frequency Density×Class Width\text{Frequency} = \text{Frequency Density} \times \text{Class Width}

💡Examples

Problem 1:

The table below shows the heights (in cm) of 20 plants. Calculate the estimated mean height.

Height (h) in cmFrequency (f)0≤h<10410≤h<20820≤h<30530≤h<403\begin{array}{|c|c|} \hline \text{Height } (h) \text{ in cm} & \text{Frequency } (f) \\ \hline 0 \le h < 10 & 4 \\ 10 \le h < 20 & 8 \\ 20 \le h < 30 & 5 \\ 30 \le h < 40 & 3 \\ \hline \end{array}

Solution:

  1. Find the midpoint (xx) for each interval:
  • 0≤h<100 \le h < 10: x=0+102=5x = \frac{0+10}{2} = 5
  • 10≤h<2010 \le h < 20: x=10+202=15x = \frac{10+20}{2} = 15
  • 20≤h<3020 \le h < 30: x=20+302=25x = \frac{20+30}{2} = 25
  • 30≤h<4030 \le h < 40: x=30+402=35x = \frac{30+40}{2} = 35
  1. Multiply frequency (ff) by midpoint (xx):
  • 4×5=204 \times 5 = 20
  • 8×15=1208 \times 15 = 120
  • 5×25=1255 \times 25 = 125
  • 3×35=1053 \times 35 = 105
  1. Sum the values:
  • ∑f=4+8+5+3=20\sum f = 4 + 8 + 5 + 3 = 20
  • ∑(f×x)=20+120+125+105=370\sum (f \times x) = 20 + 120 + 125 + 105 = 370
  1. Calculate mean: xˉ=37020=18.5 cm\bar{x} = \frac{370}{20} = 18.5 \text{ cm}

Explanation:

To estimate the mean of grouped continuous data, we use the midpoint of each class as a representative value for that group, then find the weighted average.

Problem 2:

A histogram is drawn for fixed intervals of width 55. If the frequency for the interval 10≤x<1510 \le x < 15 is 1212, what should be the height of the bar if the yy-axis represents frequency density?

Solution:

Given:

  • Frequency=12\text{Frequency} = 12
  • Class Width=15−10=5\text{Class Width} = 15 - 10 = 5

Using the formula: Frequency Density=FrequencyClass Width\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} Height=125=2.4\text{Height} = \frac{12}{5} = 2.4

Explanation:

In an extended histogram where intervals might vary (though here they are fixed), the height is determined by the frequency density to ensure the area represents the frequency correctly.