NCERT Class 9 Solutions: Statistics (Chapter 14) Exercise 14.3-Part 4 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q-9 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters of the English alphabet in the surnames was found as follows:

Frequency Distribution of Number of Letters in the Surnames
Number of lettersNumber of surnames
1 - 46
4 - 630
6 - 844
8 - 1216
12 - 204
  1. Draw a histogram to depict the given information.
  2. Write the class interval in which the maximum number of surname lies.

Solution (I) :

The problem with this data is that the class intervals are of varying width. The principle behind histograms is that the area of each bar represents the fraction of a frequency (probability) distribution within each bin (class, interval) . Therefore, as the bar in a histogram is spread across bins, it is squished in height (that is, area remains the same) . Here we choose the bin size to be 2 letters. So we divide the frequency (number of surnames names) into bins of 2 letters.

For example, number of surnames with 1 – 4 letters is 6 however since 4 – 1 = 3, this frequency is not spread across 2 letters but 3. The height in frequency representation in histogram is the proportion of the number of surnames per 2 letters interval, that is .

We calculate the histogram heights for other intervals as follows:

Height of Rectangle for Frequency Distribution of Letters of Names
Number of lettersFrequency (Number of surnames)Width of classHeight of rectangle
1 - 463
4 - 6302
6 - 8442
8 - 12164
12 - 2048

Now plot the histogram

On x-axis represent the class intervals of the number of letters and on y-axis the proportion of the number of surnames per 2 letters interval.

Chart Representing the Data

Solution (II) :

The maximum number of surnames lies in interval of 6 - 8 and it has 44 surnames.