Question

Let \( a, b \in \mathbb{R} \). Let the mean and the variance of 6 observations \(-3, 4, 7, -6, a, b\) be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:

• A. \( \frac{13}{3} \)
• B. \( \frac{16}{3} \)
• C. \( \frac{11}{3} \)
• D. \( \frac{14}{3} \)

Answer 1

The Correct Option is A

The objective is to calculate the mean deviation about the mean for six provided observations. The process is detailed in sequential steps:

1. Mean and Variance Definitions: The mean (average) of a set of observations is calculated using the formula:

\[\text{Mean} = \frac{\sum x_i}{n}\]

1. The mean of the observations \(-3, 4, 7, -6, a, b\) is given as 2:

\[\frac{-3 + 4 + 7 - 6 + a + b}{6} = 2\]

1. The equation simplifies to:

\[2 = \frac{2 + a + b}{6}\]

1. The value of \( a + b \) is determined by solving:

\[2 = \frac{2 + a + b}{6} \Rightarrow 12 = 2 + a + b \Rightarrow a + b = 10\]

1. Variance Calculation: Variance is computed using the formula:

\[\text{Variance} = \frac{\sum (x_i - \text{mean})^2}{n}\]

1. Given a variance of 23:

\[\frac{(-3-2)^2 + (4-2)^2 + (7-2)^2 + (-6-2)^2 + (a-2)^2 + (b-2)^2}{6} = 23\]

1. The squared differences from the mean are:

\[(x_i - 2)^2 = [(-3 - 2)^2 = 25, \, (4 - 2)^2 = 4, \, (7 - 2)^2 = 25,\, (-6 - 2)^2 = 64]\]

1. Substituting these values yields:

\[\frac{25 + 4 + 25 + 64 + (a-2)^2 + (b-2)^2}{6} = 23\]

1. This simplifies to:

\[\Rightarrow 118 + (a-2)^2 + (b-2)^2 = 138\]

1. Leading to:

\[\Rightarrow (a-2)^2 + (b-2)^2 = 20\]

1. Solving the System of Equations: The values of \(a\) and \(b\) are found by solving the following two equations:

\[a + b = 10 \quad \text{(equation 1)}\]

\[(a-2)^2 + (b-2)^2 = 20 \quad \text{(equation 2)}\]

1. Substituting \(b = 10 - a\) into equation 2 gives:

\[(a-2)^2 + ((10-a)-2)^2 = 20\]

1. Which further simplifies to:

\[(a-2)^2 + (8-a)^2 = 20\]

1. Expanding and solving the equation:

\[(a-2)^2 + (8-a)^2 = 20 \Rightarrow a^2 - 4a + 4 + 64 - 16a + a^2 = 20\]

\[2a^2 - 20a + 68 = 20\]

\[2a^2 - 20a + 48 = 0 \Rightarrow a^2 - 10a + 24 = 0\]

1. The quadratic equation \(a^2 - 10a + 24 = 0\) is solved for \(a\):

\[a = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm 2}{2}\]

1. Thus, \(a = 6\) or \(a = 4\). If \(a = 6\), then \(b = 4\), and if \(a = 4\), then \(b = 6\). The values for \(a\) and \(b\) are 4 and 6.
2. Mean Deviation Calculation: The mean deviation about the mean is defined as:

\[\text{Mean Deviation} = \frac{\sum |x_i - \text{mean}|}{n}\]

1. The mean deviation is computed as follows:

\[\sum |x_i - 2| = |-3-2| + |4-2| + |7-2| + |-6-2| + |a-2| + |b-2|\]

1. Using the determined values of \(a\) and \(b\):

\[|-3-2| + |4-2| + |7-2| + |-6-2| + |4-2| + |6-2|\]

1. The sum of the absolute deviations is:

\[|5| + |2| + |5| + |8| + |2| + |4| = 26\]

1. The mean deviation is therefore:

\[\text{Mean Deviation} = \frac{26}{6} = \frac{13}{3}\]