Question

A crate contains $750$ fruits. The weight of $10$ fruits from the crate taken at random are shown in the given table. Can we determine if this sample of $10$ fruits can represent the whole crate in terms of weight?
MAD of the crate $=0.5$.

• A. It may be used, as the MAD doesn't matter.
• B. No, it cannot be used to represent the crate.
• C. Yes, it can be used to represent the crate.
• D. Not enough information

Answer 1

The correct option is B No, it cannot be used to represent the crate.

Mean of the given sample $=\frac{\text{Total weight of the fruits}}{\text{Number of fruits}}=\frac{(0.8\times 1)+(1\times 2)+(1.2\times 3)+(2\times 2)+(2.4\times 2)}{1+2+3+2+2}=\frac{0.8+2+3.6+4+4.8}{10}=\frac{15.2}{10}=1.52$
MAD of the given sample $=\frac{\text{sum of all deviations}}{\text{Total number of fruits taken}}=\frac{\left|\begin{array}{c}(0.8-1.52)\times 1\end{array}\right|+(\left|\begin{array}{c}1-1.52\end{array}\right|\times 2)+(\left|\begin{array}{c}1.2-1.52\end{array}\right|\times 3)+(\left|\begin{array}{c}2-1.52\end{array}\right|\times 2)+\left(\left|\begin{array}{c}2.4-1.52\end{array}\right|\right)\times 2}{1+2+3+2+2}$
$=\frac{0.72+1.04+0.96+0.96+1.76}{10}=\frac{5.44}{10}=0.544$
Since the MAD of the sample taken does not match the MAD of the fruit crate, this sample cannot represent the given crate of fruits.