Let `bar(x)` and M.D be the mean and the mean deviation about `bar(x)` of n observations `x_(p) , i=1,` 2 . . . . . n . If each of the observation is increased by 5, then the new mean deviation about the new mean, respectively are .

Options

A. `bar(x) , M.D`
B. `bar(x) + 5, M.D`
C. `bar(x), M.D. + 5`
D. `bar(x) - 15 , m M. D. + 5`

Answer

To solve the problem step by step, we will analyze the effect of increasing each observation by 5 on the mean and the mean deviation.

### Step 1: Calculate the original mean (x̄)
The mean of n observations \( x_1, x_2, \ldots, x_n \) is given by the formula:
\[ \bar{x} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} \]

**Hint:** Remember that the mean is simply the sum of all observations divided by the number of observations.

### Step 2: Increase each observation by 5
If we increase each observation by 5, the new observations become:
\[ x_1 + 5, x_2 + 5, x_3 + 5, \ldots, x_n + 5 \]

**Hint:** Think about how adding a constant to each observation affects the overall sum.

### Step 3: Calculate the new mean (x̄')
The new mean after increasing each observation by 5 is:
\[ \bar{x}' = \frac{(x_1 + 5) + (x_2 + 5) + (x_3 + 5) + \ldots + (x_n + 5)}{n} \]
This can be simplified to:
\[ \bar{x}' = \frac{x_1 + x_2 + x_3 + \ldots + x_n + 5n}{n} = \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} + \frac{5n}{n} = \bar{x} + 5 \]

**Hint:** Notice how the new mean is simply the old mean plus the constant added to each observation.

### Step 4: Calculate the original mean deviation (M.D)
The mean deviation about the mean is defined as:
\[ \text{M.D} = \frac{|x_1 - \bar{x}| + |x_2 - \bar{x}| + |x_3 - \bar{x}| + \ldots + |x_n - \bar{x}|}{n} \]

**Hint:** The mean deviation measures the average distance of each observation from the mean.

### Step 5: Calculate the new mean deviation (M.D')
For the new observations, the mean deviation about the new mean can be calculated as:
\[ \text{M.D}' = \frac{|(x_1 + 5) - (\bar{x} + 5)| + |(x_2 + 5) - (\bar{x} + 5)| + \ldots + |(x_n + 5) - (\bar{x} + 5)|}{n} \]
This simplifies to:
\[ \text{M.D}' = \frac{|x_1 - \bar{x}| + |x_2 - \bar{x}| + \ldots + |x_n - \bar{x}|}{n} = \text{M.D} \]

**Hint:** The mean deviation remains unchanged because the constant added to each observation cancels out in the calculation.

### Conclusion
Thus, after increasing each observation by 5, the new mean is \( \bar{x} + 5 \) and the new mean deviation remains the same as the original mean deviation \( \text{M.D} \).

### Final Answer
The correct option is:
\[ \bar{x} + 5 \quad \text{and} \quad \text{M.D} \]