Question

students reported practicing violin during the last semester for 45, 38, 52, 58, and 42 hours. what is the mean absolute deviation for the number of hours students practiced the violin? round to the nearest tenth.
options: 6.4, 6.8, 32, 47

Explanation:

Step1: Find the mean

First, calculate the mean (\(\bar{x}\)) of the data set \(45, 38, 52, 58, 42\). The formula for the mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 5\) (the number of data points) and \(x_{i}\) are the data values.

$$ \bar{x}=\frac{45 + 38+52 + 58+42}{5}=\frac{235}{5} = 47 $$

Step2: Find the absolute deviations

Next, find the absolute deviation of each data point from the mean. The absolute deviation for a data point \(x_{i}\) is \(|x_{i}-\bar{x}|\).

• For \(x_1 = 45\): \(|45 - 47|=2\)
• For \(x_2 = 38\): \(|38 - 47| = 9\)
• For \(x_3 = 52\): \(|52 - 47|=5\)
• For \(x_4 = 58\): \(|58 - 47| = 11\)
• For \(x_5 = 42\): \(|42 - 47|=5\)

Step3: Find the mean of absolute deviations

Now, calculate the mean of these absolute deviations. The formula for the mean absolute deviation (MAD) is \(MAD=\frac{\sum_{i = 1}^{n}|x_{i}-\bar{x}|}{n}\).

$$ MAD=\frac{2 + 9+5 + 11+5}{5}=\frac{32}{5}=6.4 $$

Answer:

6.4