Home » Past Questions » Mathematics » \(\begin{array}{c|c} Weight(s) & 0 -10 & 10 - 20 & 20 - 30 & 40 - 50\\ \hline \text{Number of coconuts} & 10 & 27 & 19 & 6 & 2\end{array}\)Estimate the mode of the frequency distribution above....

\(\begin{array}{c|c} Weight(s) & 0 -10 & 10 - 20 & 20 - 30 & 40 - 50\\ \hline \text{Number of coconuts} & 10 & 27 & 19 & 6 & 2\end{array}\)Estimate the mode of the frequency distribution above....


Question

\(\begin{array}{c|c} Weight(s) & 0 -10 & 10 - 20 & 20 - 30 & 40 - 50\\ \hline \text{Number of coconuts} & 10 & 27 & 19 & 6 & 2\end{array}\)

Estimate the mode of the frequency distribution above.

Options

A) 13.2g

B) 15.0g

C) 16.8g

D) 17.5g

The correct answer is C.

Explanation:

Mode = a + \(\frac{(b - a)(F_m - F_b)}{2F_m - F_a - F_b}\)
= \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\)
= \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\)
= 10 + \(\frac{170}{25}\)
= 10 + 6.8
= 16.8

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Dicussion (1)

  • Mode = a + \(\frac{(b - a)(F_m - F_b)}{2F_m - F_a - F_b}\)
    = \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\)
    = \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\)
    = 10 + \(\frac{170}{25}\)
    = 10 + 6.8
    = 16.8

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