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Statistics - Graphical Representation (Histograms, Frequency Polygons, Ogives)

Grade 10ICSE

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

🔑Concepts

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A Histogram is a graphical representation of a frequency distribution using adjacent rectangles. The width represents the class interval and the height represents the frequency. For unequal class widths, heights must be adjusted using the formula: Adjusted Frequency=FrequencyClass Width×Minimum Class Width\text{Adjusted Frequency} = \frac{\text{Frequency}}{\text{Class Width}} \times \text{Minimum Class Width}

A basic histogram showing three bars of different heights representing frequencies for class intervals 0-10, 10-20, and 20-30.
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A Frequency Polygon is formed by joining the mid-points (class marks) of the tops of the histogram rectangles. To complete the polygon, the ends are joined to the mid-points of the preceding and succeeding empty classes on the x-axis.

A frequency polygon showing a line graph connecting class mid-points, starting and ending on the x-axis.
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An Ogive (Cumulative Frequency Curve) is a smooth curve drawn by plotting upper class limits against cumulative frequencies. It is used to find partition values like the Median, Q1Q_1, and Q3Q_3. To find the Median, locate the N2th\frac{N}{2}^{th} term on the y-axis, move horizontally to the curve, and then vertically down to the x-axis.

An S-shaped Ogive curve with dotted lines indicating how to find the median from the N/2 value on the y-axis.
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When class intervals are inclusive (e.g., 10-19, 20-29), they must be converted to exclusive form (9.5-19.5, 19.5-29.5) using an adjustment factor before drawing any graph. This ensures continuity on the x-axis.

📐Formulae

Class Mark(xi)=Upper Limit+Lower Limit2\text{Class Mark} (x_i) = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}

Adjustment Factor=Lower Limit of a class−Upper Limit of previous class2\text{Adjustment Factor} = \frac{\text{Lower Limit of a class} - \text{Upper Limit of previous class}}{2}

Adjusted Frequency (for unequal classes)=Frequency of the classWidth of the class×Minimum class width\text{Adjusted Frequency (for unequal classes)} = \frac{\text{Frequency of the class}}{\text{Width of the class}} \times \text{Minimum class width}

Total Frequency(N)=∑fi\text{Total Frequency} (N) = \sum f_i

Median Position=(N2)th term\text{Median Position} = \left(\frac{N}{2}\right)^{th} \text{ term}

Lower Quartile (Q1) Position=(N4)th term\text{Lower Quartile (Q1) Position} = \left(\frac{N}{4}\right)^{th} \text{ term}

Upper Quartile (Q3) Position=(3N4)th term\text{Upper Quartile (Q3) Position} = \left(\frac{3N}{4}\right)^{th} \text{ term}

Inter-quartile Range=Q3−Q1\text{Inter-quartile Range} = Q_3 - Q_1

💡Examples

Problem 1:

Draw a histogram for the following distribution and find the Mode graphically:

Class0-1010-2020-3030-4040-50
Freq51220158

Solution:

  1. Plot the class intervals 0−10,10−20,…,40−500-10, 10-20, \dots, 40-50 on the x-axis.
  2. Plot frequencies on the y-axis (Scale: 1 cm=5 units1 \text{ cm} = 5 \text{ units}).
  3. Construct rectangles for each class with heights 5,12,20,15,5, 12, 20, 15, and 88.
  4. Identify the modal class: 20−3020-30 (tallest rectangle, height =20= 20).
  5. Draw a line from the top-left corner of the 20−3020-30 rectangle to the top-left corner of the 30−4030-40 rectangle.
  6. Draw another line from the top-right corner of the 20−3020-30 rectangle to the top-right corner of the 10−2010-20 rectangle.
  7. Locate the intersection point of these two diagonal lines.
  8. Draw a perpendicular line from this intersection to the x-axis. The value on the x-axis is approximately 2626.

Mode≈26\text{Mode} \approx 26.

Explanation:

The mode is found by examining the area of highest frequency. The intersection of the diagonal lines accounts for the influence of the frequencies of classes immediately preceding and following the modal class.

Problem 2:

Given the following frequency distribution, draw an Ogive and estimate the Median:

Marks10-2020-3030-4040-5050-60
Students4915102

Solution:

  1. Calculate Cumulative Frequencies (CFCF):
    • 10−20:410-20: 4
    • 20−30:4+9=1320-30: 4 + 9 = 13
    • 30−40:13+15=2830-40: 13 + 15 = 28
    • 40−50:28+10=3840-50: 28 + 10 = 38
    • 50−60:38+2=4050-60: 38 + 2 = 40
  2. Points to plot (UpperLimit,CF)(Upper Limit, CF): (20,4),(30,13),(40,28),(50,38),(60,40)(20, 4), (30, 13), (40, 28), (50, 38), (60, 40).
  3. Include the starting point: (10,0)(10, 0).
  4. Plot these points on a graph and join them with a smooth freehand curve.
  5. Total frequency N=40N = 40. Median position =N2=402=20th= \frac{N}{2} = \frac{40}{2} = 20^{th} term.
  6. On the y-axis, find the value 2020. Draw a horizontal line to the Ogive.
  7. From the intersection point on the Ogive, drop a vertical line to the x-axis. The value on the x-axis is the Median marks.

Median≈34.6\text{Median} \approx 34.6

Explanation:

The Ogive represents the cumulative distribution. By finding the middle point of the total frequency (N/2N/2) on the y-axis, we can trace back to the x-axis to find the value below which 50%50\% of the data lies.

Problem 3:

Draw a histogram for the following frequency distribution of weights of 30 students and use it to estimate the mode:

Weight (kg)40−4545−5050−5555−6060−65No. of Students481242\begin{array}{|c|c|c|c|c|c|} \hline \text{Weight (kg)} & 40-45 & 45-50 & 50-55 & 55-60 & 60-65 \\ \hline \text{No. of Students} & 4 & 8 & 12 & 4 & 2 \\ \hline \end{array}

Identify the modal class and find the mode from the graph.

Histogram showing modal class 50-55 with intersecting lines to find the mode at 52.5.

Solution:

  1. Convert the data into a histogram where the x-axis is Weight and the y-axis is Frequency.
  2. The modal class is 50−5550-55 as it has the highest frequency (1212).
  3. To find the mode graphically: a. Join the top-right corner of the modal bar to the top-right corner of the preceding bar. b. Join the top-left corner of the modal bar to the top-left corner of the succeeding bar. c. The x-coordinate of the intersection point of these two lines is the Mode.
  4. From the graph, the Mode ≈52.5\approx 52.5 kg.

Explanation:

Mode is the value of the variable which has the maximum frequency. In a histogram, the tallest rectangle represents the modal class. The intersection of diagonals drawn from the corners of the adjacent bars provides a reliable estimate of the mode.

Problem 4:

For the following data, draw a 'less than' ogive and estimate the Lower Quartile (Q1Q_1):

Marks0−1010−2020−3030−4040−50Frequency51020105\begin{array}{|c|c|c|c|c|c|} \hline \text{Marks} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 10 & 20 & 10 & 5 \\ \hline \end{array} Total frequency N=50N = 50.

Ogive curve for marks distribution showing the projection from 12.5 on the y-axis to find Q1 on the x-axis.

Solution:

  1. Calculate Cumulative Frequencies (CF):
    • <10:5<10: 5
    • <20:15<20: 15
    • <30:35<30: 35
    • <40:45<40: 45
    • <50:50<50: 50
  2. Plot the points (10,5),(20,15),(30,35),(40,45),(50,50)(10, 5), (20, 15), (30, 35), (40, 45), (50, 50) and join them with a free-hand curve.
  3. The total frequency N=50N = 50.
  4. Position of Lower Quartile (Q1Q_1) = N4=504=12.5th\frac{N}{4} = \frac{50}{4} = 12.5^{th} term.
  5. On the y-axis, locate 12.512.5. Draw a horizontal line to meet the curve, then a vertical line down to the x-axis.
  6. Q1≈17.5Q_1 \approx 17.5 marks.

Explanation:

Quartiles divide the data into four equal parts. Q1Q_1 is the value below which 25% of the data falls. By using the cumulative frequency curve, we can interpolate this value between class boundaries.