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Use graph paper for this question.

A survey regarding height (in cm) of 60 boys belonging to class 10 of a school was conducted. The following data was recorded :

Height (in cm)No. of boys
135 - 1404
140 - 1458
145 - 15020
150 - 15514
155 - 1607
160 - 1656
165 - 1701

Taking 2 cm = height of 10 cm on one axis and 2 cm = 10 boys along the other axis, draw an ogive of the above distribution. Use the graph to estimate the following :

(i) median

(ii) lower quartile

(iii) if above 158 is considered as the tall boy of the class, find the number of boys in the class who are tall.

Measures of Central Tendency

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Answer

  1. The cumulative frequency table for the given continuous distribution is :
Height (in cm)No. of boysCumulative frequency
135 - 14044
140 - 145812
145 - 1502032
150 - 1551446
155 - 160753
160 - 165659
165 - 170160
  1. Take 2 cm along x-axis = 5 cm

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

  3. Since, scale on x-axis starts at 135, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 135.

  4. Plot the points (140, 4), (145, 12), (150, 32), (155, 46), (160, 53), (165, 59) and (170, 60) representing upper class limits and the respective cumulative frequencies.
    Also plot the point representing lower limit of the first class i.e. 135 - 140.

  5. Join these points by a freehand drawing.

A survey regarding height (in cm) of 60 boys belonging to class 10 of a school was conducted. Draw an ogive of the above distribution. Use the graph to estimate the median, lower quartile, if above 158 is considered as the tall boy of the class, find the number of boys in the class who are tall. Measures of Central Tendency, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 60.

To find the median :

Let A be the point on y-axis representing frequency = n2=602\dfrac{n}{2} = \dfrac{60}{2} = 30.

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 149.5.

Hence, the median height = 149.5 cm.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = n4=604\dfrac{n}{4} = \dfrac{60}{4} = 15.

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 146.

Hence, lower quartile = 146 cm.

(iii) Let O be the point on x-axis representing height = 158 cm.

Through O, draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 51.

No. of boys shorter than 158 cm = 51

So, no. of boys taller than 158 cm = Total boys - boys shorter than 158 cm = 60 - 51 = 9.

Hence, there are 9 tall boys in the class.

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