The sum of deviations of a set of values `x_(1),x_(2),......,x_(n)` measured from 20 is 8 and the sum of deviations of the value from 18 is 52. The value of n and the mean is

Options

A. 22,20.36
B. 25, 49.5
C. 30,50.5
D. 35,50.6

Answer

To solve the problem, we need to find the value of \( n \) and the mean of the set of values \( x_1, x_2, \ldots, x_n \) given the sum of deviations from two different points.

### Step 1: Set up the equations based on the information provided.

1. The sum of deviations from 20 is given as 8:
\[ \sum_{i=1}^{n} (x_i - 20) = 8 \]
This can be rewritten as:
\[ \sum_{i=1}^{n} x_i - 20n = 8 \]
Thus, we can express the sum of the values:
\[ \sum_{i=1}^{n} x_i = 20n + 8 \quad \text{(Equation 1)} \]

2. The sum of deviations from 18 is given as 52:
\[ \sum_{i=1}^{n} (x_i - 18) = 52 \]
This can be rewritten as:
\[ \sum_{i=1}^{n} x_i - 18n = 52 \]
Thus, we can express the sum of the values:
\[ \sum_{i=1}^{n} x_i = 18n + 52 \quad \text{(Equation 2)} \]

### Step 2: Set the two equations equal to each other.

From Equation 1 and Equation 2, we have:
\[ 20n + 8 = 18n + 52 \]

### Step 3: Solve for \( n \).

Rearranging the equation:
\[ 20n - 18n = 52 - 8 \]
\[ 2n = 44 \]
\[ n = \frac{44}{2} = 22 \]

### Step 4: Find the mean.

Now that we have \( n = 22 \), we can find the mean using the sum of the values. We can substitute \( n \) back into either Equation 1 or Equation 2. Let's use Equation 1:
\[ \sum_{i=1}^{n} x_i = 20n + 8 = 20(22) + 8 = 440 + 8 = 448 \]

Now, the mean \( \bar{x} \) is given by:
\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{448}{22} \]
Calculating this:
\[ \bar{x} = 20.36 \]

### Final Results

Thus, the value of \( n \) is \( 22 \) and the mean is \( 20.36 \).

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