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100 pupils in a school have heights as tabulated below :

Height (in cm)No. of pupils
121 - 13012
131 - 14016
141 - 15030
151 - 16020
161 - 17014
171 - 1808

Draw the ogive for the above data and from it determine the median (use graph paper).

Measures of Central Tendency

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Answer

Converting the discontinuous data into continuous data.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

=1311302=12=0.5= \dfrac{131 - 130}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5

  1. We construct the table as under :
Classes before adjustmentClasses after adjustmentNo. of pupilsCumulative frequency
121 - 130120.5 - 130.51212
131 - 140130.5 - 140.51628
141 - 150140.5 - 150.53058
151 - 160150.5 - 160.52078
161 - 170160.5 - 170.51492
171 - 180170.5 - 180.58100
  1. Take 1 cm along x-axis = 10 cm

  2. Take 1 cm along y-axis = 10 (people)

  3. Since, scale on x-axis starts at 120.5, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 120.5

  4. Plot the points (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92) and (180.5, 100) representing upper class limits and the respective cumulative frequencies.
    Also plot the point representing lower limit of the first class i.e. 120.5 - 130.5.

  5. Join these points by a freehand drawing.

100 pupils in a school have heights as tabulated below. Draw the ogive for the above data and from it determine the median (use graph paper). Measures of Central Tendency, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

The required ogive is shown in figure above.

Here, n (no. of students) = 100.

To find the median :

Let A be the point on y-axis representing frequency = n2=1002\dfrac{n}{2} = \dfrac{100}{2} = 50.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents height = 147.5 cm.

Hence, the median height = 147.5 cm.

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