EX 13.3 (Question: 6)

EX 13.3 (Question: 6) 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: Given: Total surnames = 100 Assumed mean 𝑎=8.5 Class width ℎ=3 Sum of frequencies ∑𝑓𝑖=100 Sum of 𝑓𝑖𝑢𝑖=−6 (where 𝑢𝑖 are deviations from assumed mean) Median class = 7 – 10 Cumulative frequency before median class 𝑐𝑓=36 Frequency of median class 𝑓=40 Using formula: Mean, 𝒙 ‾=𝒂+((∑1▒ 𝒇_𝒊 𝒖_𝒊)/(∑1▒ 𝒇_𝒊 ))×𝒉 Mode, =𝒍+[(𝒇_𝟏−𝒇_𝟎)/(𝟐𝒇_𝟏−𝒇_𝟎−𝒇_𝟐 )]×𝒉 Median, =𝒍+[(𝒏/𝟐−𝒄𝒇)/𝒇]×𝒉 For Mode: 𝑓1 = 40 (frequency of modal class) 𝑓0 = 30 (frequency of class before modal class) 𝑓2=16 (frequency of class after modal class) Mean 𝑥 ‾=8.5+((−6)/100)×3 =8.5−0.18 =8.32Mean = 8.32 letters Median Where: 𝑙=7 (lower limit of median class) 𝑛=100 𝑐𝑓=36 𝑓=40 ℎ=3 "Median "=7+((50−36)/40)×3 =7+14/40×3 =7+1.05=8.05Median = 8.05 letters Mode Where: 𝑙=7 (lower limit of modal class) 𝑓1=40 𝑓0=30 𝑓2=16 ℎ=3h=3 Mode =7+((40−30)/(2×40−30−16))×3 =7+(10/34)×3 =7+0.88 ="7.88"Mode = 7.88 letters

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