Question

# We are given the following marks secured by 25 students in an examination. 23, 28,30, 32, 35, 36, 36, 40, 41, 43, 44, 44, 45, 48, 49, 52, 53, 54, 56, 56, 58, 61, 62, 65, 68. (i) Arrange this data in the form of a frequency distribution taking the following class intervals. 20−29, 30−39, 40−49, 50−59 and 60−69 (ii) Draw the frequency polygon and ogive for the above data.

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Solution

## (i) The given data is arranged in the form of a frequency distribution as follows. Class Interval (Marks) Frequency (No. of Student) 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 2 5 8 6 4 $\Sigma f=25$ In order to draw a frequency polygon without a histogram, we calculate the mid-points of each of the class intervals and plot them on a graph against their respective frequencies. The curve obtained on joining the points is the frequency polygon. Marks Mid Value No. of Students 20 − 29 $\frac{20+29}{2}=24.5$ 2 30 − 39 $\frac{30+39}{2}=34.5$ 5 40 − 49 $\frac{40+49}{2}=44.5$ 8 50 − 59 $\frac{50+59}{2}=54.5$ 6 60 − 69 $\frac{60+69}{2}=64.5$ 4 Total 25 In order to draw a less than ogive, we first convert the frequency distribution as prepared above in a less than cumulative frequency distribution as follows. Less than Ogive Cumulative Frequency Less than 29 Less than 39 Less than 49 Less than 59 Less than 69 2 2 + 5 = 7 7 + 8 = 15 15 + 6 = 21 21 + 4 = 25 We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.

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