Question

The correlation coefficient between $x$and $y$ from the following data $\sum x=40$, $\sum y=50$, $\sum xy=220$, $\sum x^2=200$, $\sum y^2=262$, $n=10$is

• A. $0.89$
• B. $0.76$
• C. $0.91$
• D. $0.98$

Answer 1

The correct option is C

$0.91$

Step 1: Finding $cov(x,y)$

Given $\sum x=40$,$\sum y=50$,$\sum xy=220$, $\sum x^2=200$,$\sum y^2=262$, $n=10$

We know that $cov(x,y)=\left[\frac{\sum xy}{n}-\overline{x}.\overline{y}\right]$

Let's find the value of $\overline{x},\overline{y}$

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{40}{10} \\ =4 \end{gathered}$$

$$\begin{gathered} \overline{y}=\frac{\sum y}{n} \\ =\frac{50}{10} \\ =5 \end{gathered}$$

Substituting the values in the formula, we get

$\begin{array}{l}cov(x,y)=\left[\frac{220}{10}-(4\times 5)\right]\\ =22-20\\ =2\end{array}$

Therefore,$\begin{array}{l}cov(x,y)=2\end{array}$.

Step 2: Finding the standard deviation

The standard deviation is given by

$\begin{array}{l}{S}_x=\sqrt{\frac{\sum x^2}{n}-(\overline{x}{)}^2}\\ =\sqrt{\frac{200}{10}-(4{)}^2}\\ =\sqrt{20-16}\\ =\sqrt{4}\\ =2\end{array}$

$\begin{array}{l}{S}_y=\sqrt{\frac{\sum y^2}{n}-(\overline{y}{)}^2}\\ =\sqrt{\frac{262}{10}-(5{)}^2}\\ =\sqrt{26.2-25}\\ =\sqrt{1.2}\\ =1.0954\end{array}$

Step 3: Finding the correlation coefficient

Correlation coefficient of $x$and $y$ is

$$\begin{gathered} r=\frac{cov(x,y)}{S_x.S_y} \\ =\frac{2}{2\times 1.0954} \\ =0.91 \end{gathered}$$

Hence, option (C) is the correct answer