No. of letters 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 No. of surnames 6 30 44 16 4 Determine the class mark of each class.

Consider the given data. Class mark is the mean of the upper class limit and lower class limit of a class. Now, the given table is modified to add the column of class marks . No. of letters Class marks No. of surnames 0 - 4 0 + 4 2 = 4 2 = 2 6 4 - 8 4 + 8 2 = 12 2 = 6 30 8 - 12 8 + 12 2 = 20 2 = 10 44 12 - 16 12 + 16 2 = 14 16 16 - 20 16 + 20 2 = 18 4

No. of letters 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 No. of surnames 6 30 44 16 4 What is the upper class limit of third class?

Consider the given data. Now the third class (no.of letters) is 8 - 12 . Here upper class limit is the greatest number in the given class. So, upper class limit of 8 - 12 is 12 .

No. of letters 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 No. of surnames 6 30 44 16 4 Determine the class size.

Consider the given data. Class size is the difference between upper class limit and lower class limit of a class. Here, in the given data, all classes have the same class size. Class size of first class is 4 - 0 = 4 .

No. of letters 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 No. of surnames 6 30 44 16 4 Construct a cumulative frequency table.

Consider the given data. Cumulative frequency of a class is the sum of all frequencies upto and including the class considered. Now the cumulative frequency table is constructed as given below. No.of letters (class) No.of surnames (frequency) Cumulative frequency 0 - 4 6 6 4 - 8 30 6 + 30 = 36 8 - 12 44 36 + 44 = 80 12 - 16 16 80 + 16 = 96 16 - 20 4 96 + 4 = 100

Construct a histogram for the following data. No. of letters 1 - 4 4 - 6 6 - 8 8 - 12 12 - 20 No of surnames 6 30 44 16 4

We know, Histogram is a graphical representation of frequency distribution in form of rectangles with bases as class-intervals and heights as corresponding frequencies. In histogram, there is no gap between any rectangles. Here the class sizes are not equal for all class intervals. So we have to adjust the frequencies such that the class size of each class interval is 2 . Now the adjusted frequency distribution table is shown below. No.of letters(class interval) No. of surnames(frequency) Class size Adjusted frequency 1 - 4 6 3 6 3 × 2 = 4 4 - 6 30 2 30 2 × 2 = 30 6 - 8 44 2 44 2 × 2 = 44 8 - 12 16 4 16 4 × 2 = 8 12 - 20 4 8 4 8 × 2 = 1 Draw the histogram by representing no.of letters (class-interval) on X-axis and adjusted frequency on Y-axis , as shown below.