Sum of the squares of deviation from the mean of x series is 136 and that of y series is 13. Sum of the product of the deviations of x and y series from their respective means is 122. Find the Pearson's coefficient of correlation.

Answer

To find the Pearson's coefficient of correlation (denoted as \( r \)), we can use the formula:

\[ r = \frac{\sum (X - \bar{X})(Y - \bar{Y})}{\sqrt{\sum (X - \bar{X})^2 \sum (Y - \bar{Y})^2}} \]

### Step 1: Identify the given values
From the problem, we have:
- \(\sum (X - \bar{X})^2 = 136\)
- \(\sum (Y - \bar{Y})^2 = 13\)
- \(\sum (X - \bar{X})(Y - \bar{Y}) = 122\)

### Step 2: Substitute the values into the formula
We substitute the values into the formula for \( r \):

\[ r = \frac{122}{\sqrt{136 \times 13}} \]

### Step 3: Calculate the denominator
First, we calculate \( 136 \times 13 \):

\[ 136 \times 13 = 1768 \]

Now, we find the square root of \( 1768 \):

\[ \sqrt{1768} \approx 42.048 \]

### Step 4: Calculate \( r \)
Now we substitute back into the equation for \( r \):

\[ r = \frac{122}{42.048} \approx 2.901 \]

### Final Answer
Thus, the Pearson's coefficient of correlation is approximately:

\[ r \approx 2.901 \]