Questions: Question 10 1 pts Suppose that the last four months of sales were 8, 10, 15, and 9 units, respectively. Suppose further that the last four forecasts were 5, 6, 11, and 12 units, respectively. What is the Mean Absolute Deviation (MAD) of these forecasts? 10.05 3.50 2.00 0.50 8.00

Question 10 1 pts

Suppose that the last four months of sales were 8, 10, 15, and 9 units, respectively. Suppose further that the last four forecasts were 5, 6, 11, and 12 units, respectively. What is the Mean Absolute Deviation (MAD) of these forecasts?
10.05
3.50
2.00
0.50
8.00
Transcript text: Question 10 1 pts Suppose that the last four months of sales were $8,10,15$, and 9 units, respectively. Suppose further that the last four forecasts were $5,6,11$, and 12 units, respectively. What is the Mean Absolute Deviation (MAD) of these forecasts? 10.05 3.50 2.00 0.50 8.00
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Solution

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Solution Steps

Step 1: Calculate Absolute Deviations

To find the Mean Absolute Deviation (MAD), we first calculate the absolute deviations between the actual sales and the forecasts. The absolute deviations are given by:

\[ | \text{sales}_i - \text{forecasts}_i | \]

For the provided data:

  • Sales: \(8, 10, 15, 9\)
  • Forecasts: \(5, 6, 11, 12\)

Calculating the absolute deviations:

  • For \(i=1\): \( |8 - 5| = 3 \)
  • For \(i=2\): \( |10 - 6| = 4 \)
  • For \(i=3\): \( |15 - 11| = 4 \)
  • For \(i=4\): \( |9 - 12| = 3 \)

Thus, the absolute deviations are: \[ [3, 4, 4, 3] \]

Step 2: Calculate the Mean of Absolute Deviations

Next, we calculate the mean of these absolute deviations. The mean \( \mu \) is calculated as follows:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} \]

Where \( N \) is the number of observations (in this case, \( N = 4 \)) and \( x_i \) are the absolute deviations. Therefore:

\[ \mu = \frac{3 + 4 + 4 + 3}{4} = \frac{14}{4} = 3.5 \]

Final Answer

The Mean Absolute Deviation (MAD) of the forecasts is:

\[ \boxed{3.5} \]

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