The demand and forecast of an item for five months are given in the table.
\[ \begin{array}{|c|c|c|} \hline \text{Month} & \text{Demand} & \text{Forecast} \\ \hline \text{April} & 225 & 200 \\ \text{May} & 220 & 240 \\ \text{June} & 285 & 300 \\ \text{July} & 290 & 270 \\ \text{August} & 250 & 230 \\ \hline \end{array} \]
The Mean Absolute Percent Error (MAPE) in the forecast is \(\underline{\hspace{1cm}}\) % (round off to two decimal places).
The Mean Absolute Percent Error (MAPE) is given by: \[ \text{MAPE} = \frac{1}{n} \sum_{i=1}^{n} \left| \frac{\text{Actual}_i - \text{Forecast}_i}{\text{Actual}_i} \right| \times 100 \] For the given data: \[ \text{MAPE} = \frac{1}{5} \left( \left| \frac{225-200}{225} \right| + \left| \frac{220-240}{220} \right| + \left| \frac{285-300}{285} \right| + \left| \frac{290-270}{290} \right| + \left| \frac{250-230}{250} \right| \right) \times 100. \] \[ = \frac{1}{5} \left( 0.1111 + 0.0909 + 0.0526 + 0.0689 + 0.0800 \right) \times 100 = \frac{1}{5} \times 0.4035 \times 100 = 8.07%. \] Thus, the MAPE is: \[ \boxed{6.00 \, \text{to} \, 10.00 \, \%}. \]
Considering the actual demand and the forecast for a product given in the table below, the mean forecast error and the mean absolute deviation, respectively, are:

P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?