Answered

Find the value of [tex]r[/tex] from the following data.

\begin{tabular}{|l|c|c|}
\hline
Particulars & [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
Average & 60 & 95 \\
\hline
\begin{tabular}{l}
The sum of squares of deviations \\
taken from their mean
\end{tabular} & 920 & 1050 \\
\hline
\begin{tabular}{l}
The sum of product of deviations \\
taken from their mean
\end{tabular} & \multicolumn{2}{|c|}{-545} \\
\hline
\end{tabular}



Answer :

Let's walk through the steps to find the value of the correlation coefficient \( r \) from the given data.

### Step 1: Calculate the Standard Deviations
To find the correlation coefficient, we first need to calculate the standard deviations of \( x \) and \( y \).

Given:
- Average \( x \) = 60
- Average \( y \) = 95
- Sum of squares of deviations for \( x \) = 920
- Sum of squares of deviations for \( y \) = 1050
- The sum of product of deviations = -545

We assume the sample size \( n \) is such that \( n-1 = 10 \), meaning \( n = 11 \). With \( n-1 \) as our value (which is typically used in the calculation of standard deviation for samples), we compute the standard deviations.

The formula for the standard deviation of \( x \) is:
[tex]\[ \sigma_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \][/tex]

For \( x \):
[tex]\[ \sigma_x = \sqrt{\frac{920}{10}} \][/tex]
[tex]\[ \sigma_x = \sqrt{92} \][/tex]
[tex]\[ \sigma_x \approx 9.59 \][/tex]

For \( y \):
[tex]\[ \sigma_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n-1}} \][/tex]
[tex]\[ \sigma_y = \sqrt{\frac{1050}{10}} \][/tex]
[tex]\[ \sigma_y = \sqrt{105} \][/tex]
[tex]\[ \sigma_y \approx 10.25 \][/tex]

### Step 2: Calculate the Correlation Coefficient \( r \)

The formula for the correlation coefficient \( r \) is:
[tex]\[ r = \frac{\sum((x_i - \bar{x})(y_i - \bar{y}))}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} \][/tex]

Given the sum of product of deviations:
[tex]\[ \sum ((x_i - \bar{x})(y_i - \bar{y})) = -545 \][/tex]

We use this sum of product of deviations along with the standard deviations calculated above:
[tex]\[ r = \frac{\sum((x_i - \bar{x})(y_i - \bar{y}))}{(n-1) \sigma_x \sigma_y} \][/tex]

Substitute the values:
[tex]\[ r = \frac{-545}{10 \times 9.59 \times 10.25} \][/tex]

Calculate the denominator:
[tex]\[ 9.59 \times 10.25 = 98.2975 \][/tex]
[tex]\[ 10 \times 98.2975 = 982.975 \][/tex]

Now, compute \( r \):
[tex]\[ r = \frac{-545}{982.975} \][/tex]
[tex]\[ r \approx -0.5545 \][/tex]

### Conclusion
Therefore, the correlation coefficient [tex]\( r \approx -0.5545 \)[/tex].