Let `barx` be the mean of `x_(1),x_(2),x_(3),……..,x_(n)`. If `x_(i)=a+cy_(i)` for some constants a and c, then what will be the mean of `y_(1),y_(2),y_(3),……..,y_(n)`?

Options

A. `a+cbarx`
B. `a-(1)/(c)barx`
C. `(1)/(c)barx-a`
D. `(bar(x)-a)/(c)`

Explanation

To find the mean of \( y_1, y_2, y_3, \ldots, y_n \) given that \( x_i = a + c y_i \) for some constants \( a \) and \( c \), we can follow these steps:

### Step 1: Write the expression for the mean of \( x_i \)

The mean of \( x_1, x_2, \ldots, x_n \) is given by:

\[ \bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n} \]

### Step 2: Substitute the expression for \( x_i \)

Since \( x_i = a + c y_i \), we can substitute this into the mean:

\[ \bar{x} = \frac{(a + c y_1) + (a + c y_2) + \ldots + (a + c y_n)}{n} \]

### Step 3: Simplify the expression

Now, let's simplify the numerator:

\[ \bar{x} = \frac{(a + a + a + \ldots + a) + (c y_1 + c y_2 + \ldots + c y_n)}{n} \]

There are \( n \) terms of \( a \), so this becomes:

\[ \bar{x} = \frac{n a + c (y_1 + y_2 + \ldots + y_n)}{n} \]

### Step 4: Separate the terms

Now we can separate the terms in the fraction:

\[ \bar{x} = a + \frac{c (y_1 + y_2 + \ldots + y_n)}{n} \]

### Step 5: Express the mean of \( y_i \)

Let \( \bar{y} \) be the mean of \( y_1, y_2, \ldots, y_n \):

\[ \bar{y} = \frac{y_1 + y_2 + \ldots + y_n}{n} \]

Substituting this into our equation gives:

\[ \bar{x} = a + c \bar{y} \]

### Step 6: Solve for \( \bar{y} \)

Now, we can isolate \( \bar{y} \):

\[ c \bar{y} = \bar{x} - a \]

Dividing both sides by \( c \) (assuming \( c \neq 0 \)) gives:

\[ \bar{y} = \frac{\bar{x} - a}{c} \]

### Final Result

Thus, the mean of \( y_1, y_2, \ldots, y_n \) is:

\[ \bar{y} = \frac{\bar{x} - a}{c} \]