If `sum_(i=1)^(18)(x_(i) - 8) = 9` and `sum_(i=1)^(18)(x_(i) - 8)^(2) = 45` then find the standard deviation of `x_(1), x_(2),….x_(18)`

Answer

To find the standard deviation of the values \( x_1, x_2, \ldots, x_{18} \) given the conditions, we can follow these steps:

### Step 1: Calculate the Mean of \( x_i - 8 \)

We are given:
\[ \sum_{i=1}^{18} (x_i - 8) = 9 \]
To find the mean, we divide the sum by the number of terms (which is 18):
\[ \text{Mean} = \frac{\sum_{i=1}^{18} (x_i - 8)}{18} = \frac{9}{18} = \frac{1}{2} \]

### Step 2: Calculate the Mean of \( (x_i - 8)^2 \)

We are also given:
\[ \sum_{i=1}^{18} (x_i - 8)^2 = 45 \]
To find the mean of the squares, we divide by the number of terms:
\[ \text{Mean of squares} = \frac{\sum_{i=1}^{18} (x_i - 8)^2}{18} = \frac{45}{18} = \frac{5}{2} \]

### Step 3: Calculate the Variance

The variance is calculated using the formula:
\[ \text{Variance} = \text{Mean of squares} - (\text{Mean})^2 \]
Substituting the values we found:
\[ \text{Variance} = \frac{5}{2} - \left(\frac{1}{2}\right)^2 \]
Calculating \( \left(\frac{1}{2}\right)^2 \):
\[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Now substituting back into the variance formula:
\[ \text{Variance} = \frac{5}{2} - \frac{1}{4} \]
To perform the subtraction, we convert \( \frac{5}{2} \) to have a common denominator of 4:
\[ \frac{5}{2} = \frac{10}{4} \]
Now we can subtract:
\[ \text{Variance} = \frac{10}{4} - \frac{1}{4} = \frac{9}{4} \]

### Step 4: Calculate the Standard Deviation

The standard deviation is the square root of the variance:
\[ \text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \]

Thus, the standard deviation of \( x_1, x_2, \ldots, x_{18} \) is:
\[ \boxed{\frac{3}{2}} \]