The deriations from the mean of a set of numbers are \( (k-13)^{2},(k+7) \); \( -2, k \) and \( (k+2)^{2} \), Where \( k \) is a constant. Find the value of \( k \) if the mean deviation

To find the value of \( k \) given the deviations from the mean, we need to set up the equation for the mean deviation. The mean deviation is the average of the absolute deviations from the mean. The deviations from the mean are given as: 1. \( (k-13)^{2} \) 2. \( (k+7) \) 3. \( -2 \) 4. \( k \) 5. \( (k+2)^{2} \) Let's denote the deviations as \( d_1, d_2, d_3, d_4, d_5 \): - \( d_1 = (k-13)^{2} \) - \( d_2 = (k+7) \) - \( d_3 = -2 \) - \( d_4 = k \) - \( d_5 = (k+2)^{2} \) The mean deviation \( M \) can be calculated as: \[ M = \frac{1}{n} \sum_{i=1}^{n} |d_i| \] where \( n \) is the number of deviations. In this case, \( n = 5 \). Therefore, we need to calculate: \[ M = \frac{1}{5} \left( |(k-13)^{2}| + |(k+7)| + |-2| + |k| + |(k+2)^{2}| \right) \] Since all terms are non-negative, we can drop the absolute value signs: \[ M = \frac{1}{5} \left( (k-13)^{2} + (k+7) + 2 + k + (k+2)^{2} \right) \] Now, let's simplify this expression step by step. 1. Calculate \( (k-13)^{2} \): \[ (k-13)^{2} = k^{2} - 26k + 169 \] 2. Calculate \( (k+2)^{2} \): \[ (k+2)^{2} = k^{2} + 4k + 4 \] 3. Combine all terms: \[ M = \frac{1}{5} \left( (k^{2} - 26k + 169) + (k + 7) + 2 + k + (k^{2} + 4k + 4) \right) \] Combine like terms: \[ = \frac{1}{5} \left( 2k^{2} - 20k + 182 \right) \] Now, we can set the mean deviation \( M \) equal to a specific value if provided, or we can solve for \( k \) if we have additional conditions. However, since the problem does not specify a value for the mean deviation, we will need to find \( k \) such that the mean deviation is minimized or equals zero. Let's assume we want to find \( k \) such that the mean deviation is minimized. We can set the derivative of \( M \) with respect to \( k \) to zero and solve for \( k \). Let's calculate the derivative of \( M \) and set it to zero. The expression for \( M \) is: \[ M = \frac{1}{5} \left( 2k^{2} - 20k + 182 \right) \] Taking the derivative: \[ \frac{dM}{dk} = \frac{1}{5} (4k - 20) \] Setting the derivative to zero: \[ 4k - 20 = 0 \] Solving for \( k \): \[ 4k = 20 \implies k = 5 \] Now, let's confirm that this value of \( k \) is indeed a minimum by checking the second derivative: \[ \frac{d^2M}{dk^2} = \frac{1}{5} \cdot 4 = \frac{4}{5} > 0 \] Since the second derivative is positive, \( k = 5 \) is a minimum. Thus, the value of \( k \) is: \[ \boxed{5} \]