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Mathematics

Construct a combined histogram and frequency polygon for the following distribution:

ClassesFrequency
91 - 10016
101 - 11028
111 - 12044
121 - 13020
131 - 14032
141 - 15012
151 - 1604

Statistics

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Answer

The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,

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

= 1011002=12\dfrac{101 - 100}{2} = \dfrac{1}{2}

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for given data is :

Classes before adjustmentClasses after adjustmentClass markFrequency
91 - 10090.5 - 100.595.516
101 - 110100.5 - 110.5105.528
111 - 120110.5 - 120.5115.544
121 - 130120.5 - 130.5125.520
131 - 140130.5 - 140.5135.532
141 - 150140.5 - 150.5145.512
151 - 160150.5 - 160.5155.54

Steps of construction of histogram :

  1. Since, the scale on x-axis starts at 80.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 80.5

  2. Take 1 cm along x-axis = 10 units.

  3. Take 1 cm along y-axis = 5 units.

  4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

  1. Mark the mid-points of upper bases of rectangles of the histogram.

  2. Join the consecutive mid-points by line segments.

  3. Join the first end point with the mid-point of class 80.5 - 90.5 with zero frequency, and join the other end point with the mid-point of class 160.5 - 170.5 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

Construct a combined histogram and frequency polygon for the following distribution. Statistics, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

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