If a variate `x` is expressed as a linear function of two variates `u` and `v` in the form `x=au+bv`, then mean `bar(x)` of `x` is

Options

A. a) `a overline(U)-b overline(V)`
B. b) `overline (U)+overline(V)`
C. c) `b overline(U)+a overline(V)`
D. d) None of these

Explanation

To find the mean \(\bar{x}\) of the variate \(x\) expressed as a linear function of two variates \(u\) and \(v\) in the form \(x = au + bv\), we can follow these steps:

### Step 1: Write the expression for \(x\)
Given that \(x\) is expressed as:
\[ x = au + bv \]
where \(a\) and \(b\) are constants.

### Step 2: Define the mean of \(x\)
The mean \(\bar{x}\) of the variate \(x\) is defined as:
\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]
where \(n\) is the number of observations and \(x_i\) represents the individual values of \(x\).

### Step 3: Substitute the expression for \(x\)
Substituting the expression for \(x\) into the formula for the mean, we have:
\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} (au_i + bv_i) \]

### Step 4: Distribute the summation
Using the properties of summation, we can separate the terms:
\[ \bar{x} = \frac{1}{n} \left( \sum_{i=1}^{n} au_i + \sum_{i=1}^{n} bv_i \right) \]
Since \(a\) and \(b\) are constants, we can factor them out:
\[ \bar{x} = \frac{1}{n} \left( a \sum_{i=1}^{n} u_i + b \sum_{i=1}^{n} v_i \right) \]

### Step 5: Simplify the expression
Now, we can rewrite the sums as:
\[ \bar{x} = a \left( \frac{1}{n} \sum_{i=1}^{n} u_i \right) + b \left( \frac{1}{n} \sum_{i=1}^{n} v_i \right) \]
This can be expressed in terms of the means of \(u\) and \(v\):
\[ \bar{x} = a \bar{u} + b \bar{v} \]
where \(\bar{u}\) is the mean of \(u\) and \(\bar{v}\) is the mean of \(v\).

### Final Result
Thus, the mean \(\bar{x}\) of the variate \(x\) is given by:
\[ \bar{x} = a \bar{u} + b \bar{v} \]

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