If for a distribution `sum(x-7)=6` and `sum(x-7)^(2)=78` and the total number of items is 12 then mean and standard deviation are

Options

A. `barx=7.5,sigma=2.5`
B. `barx=7,sigma=2.5`
C. `barx=7.5,sigma=2`
D. `barx=7,sigma=2`

Answer

To solve the problem, we need to find the mean and standard deviation given the following information:

1. \( \sum (x - 7) = 6 \)
2. \( \sum (x - 7)^2 = 78 \)
3. Total number of items \( n = 12 \)

### Step 1: Calculate the Mean

The mean \( \bar{x} \) can be calculated using the formula:

\[ \bar{x} = a + \frac{\sum (x - a)}{n} \]

where \( a \) is the assumed mean. In this case, \( a = 7 \).

Substituting the values into the formula:

\[ \bar{x} = 7 + \frac{6}{12} \]

Calculating the fraction:

\[ \frac{6}{12} = 0.5 \]

Thus,

\[ \bar{x} = 7 + 0.5 = 7.5 \]

### Step 2: Calculate the Standard Deviation

The standard deviation \( \sigma \) can be calculated using the formula:

\[ \sigma = \sqrt{\frac{\sum (x - a)^2}{n} - \left(\frac{\sum (x - a)}{n}\right)^2} \]

Substituting the known values:

\[ \sigma = \sqrt{\frac{78}{12} - \left(\frac{6}{12}\right)^2} \]

Calculating each part:

1. Calculate \( \frac{78}{12} \):

\[ \frac{78}{12} = 6.5 \]

2. Calculate \( \left(\frac{6}{12}\right)^2 \):

\[ \left(\frac{6}{12}\right)^2 = \left(0.5\right)^2 = 0.25 \]

Now substituting these values back into the formula:

\[ \sigma = \sqrt{6.5 - 0.25} \]

Calculating the subtraction:

\[ 6.5 - 0.25 = 6.25 \]

Finally, take the square root:

\[ \sigma = \sqrt{6.25} = 2.5 \]

### Final Answers

Thus, the mean and standard deviation are:

- Mean \( \bar{x} = 7.5 \)
- Standard Deviation \( \sigma = 2.5 \)

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