Question

# $$100$$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters$$1-4$$$$4-7$$$$7-10$$$$10-13$$$$13-16$$$$16-19$$Number of surnames$$6$$$$30$$$$40$$$$16$$$$4$$$$4$$Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution

## Let us prepare the following table to compute the median :Number of letters  Number of surnames (Frequency) Cumulative frequency $$1-4$$$$6$$ $$6$$  $$4-7$$$$30$$ $$36$$  $$7-10$$$$40$$ $$76$$  $$10-13$$$$16$$ $$92$$  $$13-16$$$$4$$ $$96$$  $$16-19$$$$4$$ $$100=n$$ We have, $$n = 100$$$$\Rightarrow \dfrac n2 = 50$$The cumulative frequency just greater than $$\dfrac n2$$ is $$76$$ and the corresponding class is $$7 – 10$$. Thus, $$7 – 10$$ is the median class such that$$\dfrac n2 = 50, l = 7, f = 40, cf = 36$$ and $$h=3$$Substitute these values in the formulaMedian, $$M = l+\left(\dfrac{\dfrac n2 - cf}{f}\right)\times h$$$$M = 7+\left(\dfrac{50-36}{40}\right)\times 3$$$$M = 7+\dfrac{14}{40}\times3 = 7 + 1.05 = 8.05$$Now, calculation of mean: Number of letters Mid-Point $$(x_i)$$Frequency $$(f_i)$$ $$f_ix_i$$ $$1-4$$$$2.5$$$$6$$ $$15$$  $$4-7$$$$5.5$$ $$30$$ $$165$$  $$7-10$$$$8.5$$ $$40$$ $$340$$  $$10-13$$$$11.5$$ $$16$$ $$184$$  $$13-16$$$$14.5$$ $$4$$ $$58$$  $$16-19$$$$17.5$$ $$4$$ $$70$$  Total $$100$$ $$832$$ Therefore, Mean, $$\bar x = \dfrac{\sum f_ix_i}{\sum f_i} = \dfrac{832}{100} = 8.32$$Calculation ofMode:The class $$7 – 10$$ has the maximum frequency therefore, this is the modal class.Here, $$l = 7, h = 3, f_1 = 40, f_0 = 30$$ and $$f_2 = 16$$Now, let us substitute these values in the formulaMode $$= l+ \left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$$$= 7+\dfrac{40-30}{80-30-16}\times3$$$$= 7 + \dfrac{10}{34}\times 3 = 7+0.88 = 7.88$$Hence, median $$= 8.05$$, mean $$= 8.32$$ and mode $$= 7.88$$Mathematics

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