heart rate record\nminutes\theart rate (bpm)\n7\t121\n8\t136\n7\t122\n4\t96\n3\t90\n6\t120\n7\t118\n4\t98\n2\…

Explanation:

Step1: Calculate the means

Let $x$ be the minutes and $y$ be the heart - rate. First, find the mean of $x$ values: $\bar{x}=\frac{7 + 8+7+4+3+6+7+4+2+5+7+8}{12}=\frac{70}{12}\approx5.83$ First, find the mean of $y$ values: $\bar{y}=\frac{121 + 136+122+96+90+120+118+98+84+110+118+128}{12}=\frac{1341}{12}=111.75$

Step2: Calculate the numerator and denominator components

Calculate $(x_i-\bar{x})(y_i - \bar{y})$, $(x_i-\bar{x})^2$ and $(y_i-\bar{y})^2$ for each $i$ from $1$ to $12$. Sum them up. Let $S_{xy}=\sum_{i = 1}^{12}(x_i-\bar{x})(y_i - \bar{y})$, $S_{xx}=\sum_{i = 1}^{12}(x_i-\bar{x})^2$ and $S_{yy}=\sum_{i = 1}^{12}(y_i-\bar{y})^2$.

$S_{xy}=(7 - 5.83)(121-111.75)+(8 - 5.83)(136 - 111.75)+\cdots+(8 - 5.83)(128 - 111.75)$ $S_{xx}=(7 - 5.83)^2+(8 - 5.83)^2+\cdots+(8 - 5.83)^2$ $S_{yy}=(121-111.75)^2+(136 - 111.75)^2+\cdots+(128 - 111.75)^2$

After calculation: $S_{xy}=197.5833$ $S_{xx}=41.6667$ $S_{yy}=1393.25$

Step3: Calculate the correlation coefficient

The correlation coefficient $r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$ $r=\frac{197.5833}{\sqrt{41.6667\times1393.25}}=\frac{197.5833}{\sqrt{58068.75833}}\approx\frac{197.5833}{240.9746}\approx0.82$

Answer:

$0.82$