Q. 225.0( 2 Votes )

# The following tab

Answer :

For equal class intervals, we will solve by finding mid points of these classes using direct method. We have got

Σfi = 47 and Σfixi = 1516

mean is given by   Thus, mean is 32.26.

To find median, Assume

Σfi = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and Cf = cumulative frequency

Lets form a table. We have got

So, N = 47

N/2 = 47/2 = 23.5

The cumulative frequency just greater than (N/2 = ) 23.5 is 29, so the corresponding median class is 30.5 - 35.5 and accordingly we get Cf = 15(cumulative frequency before the median class).

Now, since median class is 30.5 - 35.5.

l = 30.5, h = 5, f = 14, N/2 = 23.5 and Cf = 15

Median is given by,  = 30 + 3.03

= 33.03

Thus, median is 33.03.

Since, we have got mean = 32.26 and median = 33.03

Applying the empirical formula,

Mode = 3(Median) – 2(Mean)

Mode = 3(33.03) – 2(32.26)

Mode = 99.09 – 64.52 = 34.57

Mean = 32.26, Median = 33.03 and Mode = 34.57

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