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.

Following are the marks of a group of 92 students in a test of reading ability: Marks 50 - 52 47 - 49 44 - 46 41 - 43 38 - 40 35 - 37 32 - 34 No of students 04 10 15 18 20 12 13 Draw a frequency polygon for the above data.

The given data is a discontinuous frequency distribution, which is converted to continuous frequency distribution as shown below. Subtract 0 . 5 from the lower limit and add 0 . 5 to upper limit of each interval. For the interval 50 - 52 , the new class interval is 50 - 0 . 5 = 49 . 5 52 + 0 . 5 = 52 . 5 Follow the similar method for other intervals as well. The new frequency distribution table is given below. Marks(clas-interval) Class marks No of students 49 . 5 - 52 . 5 51 04 46 . 5 - 49 . 5 48 10 43 . 5 - 46 . 5 45 15 40 . 5 - 43 . 5 42 18 37 . 5 - 40 . 5 39 20 34 . 5 - 37 . 5 36 12 31 . 5 - 34 . 5 32 13 Draw the frequency polygon by representing the class marks on X-axis and number of students (frequency) on Y-axis.

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MATHEMATICS>Chapter 14 - Statistics>TEST YOURSELF

A K Upadhyay and S Upadhyay Solutions for Chapter: Statistics, Exercise 4: TEST YOURSELF

Author:A K Upadhyay & S Upadhyay

A K Upadhyay Mathematics Solutions for Exercise - A K Upadhyay and S Upadhyay Solutions for Chapter: Statistics, Exercise 4: TEST YOURSELF

Attempt the practice questions on Chapter 14: Statistics, Exercise 4: TEST YOURSELF with hints and solutions to strengthen your understanding. MATHEMATICS solutions are prepared by Experienced Embibe Experts.