  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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