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

A) 13.2g

B) 15.0g

C) 16.8g

D) 17.5g

### 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

## 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