Let `x_1 ,x_2 , x_3 ………x_k` be k observations and `w_i = ax_i +b` for I = 1,2,3……. K , where a and b are constants . If mean of `x_(i)` is 52 and their standard deviation is 12 and mean of `w_i` is 60 and their standard deviation is 15, then the value of a and b should be 15, then the value of a and b should be

Options

A. `a=1.25, b = -5`
B. `a=-1.25,b=5`
C. `a=2.5,b=-5`
D. `a=2.5,b=5`

Answer

To solve the problem, we need to find the values of constants \( a \) and \( b \) given the relationships between the observations \( x_i \) and \( w_i \).

### Step 1: Understand the relationships
We know that:
- \( w_i = ax_i + b \)
- The mean of \( x_i \) is \( 52 \)
- The standard deviation of \( x_i \) is \( 12 \)
- The mean of \( w_i \) is \( 60 \)
- The standard deviation of \( w_i \) is \( 15 \)

### Step 2: Relate the means
The mean of \( w_i \) can be expressed in terms of the mean of \( x_i \):
\[ \text{Mean of } w_i = a \cdot \text{Mean of } x_i + b \]
Substituting the known values:
\[ 60 = a \cdot 52 + b \quad \text{(Equation 1)} \]

### Step 3: Relate the standard deviations
The standard deviation of \( w_i \) is given by:
\[ \text{Standard deviation of } w_i = |a| \cdot \text{Standard deviation of } x_i \]
Substituting the known values:
\[ 15 = |a| \cdot 12 \]
Solving for \( a \):
\[ |a| = \frac{15}{12} = 1.25 \]
Since \( a \) is a constant, we can take \( a = 1.25 \).

### Step 4: Substitute \( a \) back into Equation 1
Now we substitute \( a = 1.25 \) into Equation 1:
\[ 60 = 1.25 \cdot 52 + b \]
Calculating \( 1.25 \cdot 52 \):
\[ 1.25 \cdot 52 = 65 \]
Now substituting back:
\[ 60 = 65 + b \]
Solving for \( b \):
\[ b = 60 - 65 = -5 \]

### Conclusion
The values of \( a \) and \( b \) are:
\[ a = 1.25, \quad b = -5 \]