If for a distribution `sum(x-5)=3,sum(x-5)^(2)=43` and the total number of terms is 18 then mean and variance are

Options

A. `barx=(93)/(18),sigma^(2)=(85)/(36)`
B. `barx=(95)/(18),sigma^(2)=(85)/(36)`
C. `barx=(31)/(6),sigma^(2)=(85)/(36)`
D. `barx=(95)/(18),sigma^(2)=(83)/(36)`

Answer

To solve the problem, we need to find the mean and variance of the given distribution using the provided information.

### Step 1: Identify the given values
We are given:
- \( n = 18 \) (total number of terms)
- \( \sum (x - 5) = 3 \)
- \( \sum (x - 5)^2 = 43 \)

### Step 2: Calculate the mean
The mean \( \bar{x} \) can be calculated using the formula:
\[ \bar{x} = a + \frac{\sum (x - a)}{n} \]
Here, we can take \( a = 5 \).

Substituting the values:
\[ \bar{x} = 5 + \frac{\sum (x - 5)}{n} = 5 + \frac{3}{18} \]
Calculating \( \frac{3}{18} \):
\[ \frac{3}{18} = \frac{1}{6} \]
So,
\[ \bar{x} = 5 + \frac{1}{6} = \frac{30}{6} + \frac{1}{6} = \frac{31}{6} \]

### Step 3: Calculate the variance
The variance \( \sigma^2 \) can be calculated using the formula:
\[ \sigma^2 = \frac{\sum (x - 5)^2}{n} - \left( \frac{\sum (x - 5)}{n} \right)^2 \]
Substituting the values:
\[ \sigma^2 = \frac{43}{18} - \left( \frac{3}{18} \right)^2 \]
Calculating \( \left( \frac{3}{18} \right)^2 \):
\[ \left( \frac{3}{18} \right)^2 = \frac{9}{324} = \frac{1}{36} \]
Now substituting this back into the variance formula:
\[ \sigma^2 = \frac{43}{18} - \frac{1}{36} \]
To subtract these fractions, we need a common denominator. The least common multiple of 18 and 36 is 36. Thus, we convert \( \frac{43}{18} \) to have a denominator of 36:
\[ \frac{43}{18} = \frac{43 \times 2}{18 \times 2} = \frac{86}{36} \]
Now substituting:
\[ \sigma^2 = \frac{86}{36} - \frac{1}{36} = \frac{86 - 1}{36} = \frac{85}{36} \]

### Final Answers:
- Mean \( \bar{x} = \frac{31}{6} \)
- Variance \( \sigma^2 = \frac{85}{36} \)