91Ó°ÊÓ

Minute heart rate (heart rate), the number of heartbeats obtained within one minute.

Given Data

Consider the given data as below.

Counting the number of beats, t = 30 s

Number of beats, N = 40 beats

Uncertainty in beats,$\mathrm{δ\pm ·}=1$

Uncertainty in time =$\mathrm{δ³Ù}=0.5\mathrm{s}$

The expression for the number of beats per minute is

$\mathrm{K}=\raisebox{1ex}{$\mathrm{N}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.$

Substitute known values in the above equation.

$$\begin{gathered} \mathrm{K}=\frac{40\mathrm{beats}}{30\sec }×\frac{60\sec }{1\min } \\ =80\mathrm{beats}/\min \end{gathered}$$

The maximum number of beats is

$$\begin{gathered} {\mathrm{N}}_{\max }=\mathrm{N}+\mathrm{δ\pm ·} \\ =40+1 \\ =41\mathrm{beats} \end{gathered}$$

The minimum time is,-

$$\begin{gathered} {\mathrm{t}}_{\min }=\mathrm{t}-\mathrm{δ³Ù} \\ =30-0.5 \\ =29.5\sec \end{gathered}$$

The maximum number of beats per minute is,-

$$\begin{gathered} {\mathrm{K}}_{\max }=\frac{{\mathrm{N}}_{\max }}{{\mathrm{t}}_{\min }} \\ =\frac{41\mathrm{beats}}{29.5\sec }×\frac{60\sec }{1\min } \\ =83.4\mathrm{beats}/\min \end{gathered}$$

The minimum number of beats is,-

$$\begin{gathered} {\mathrm{N}}_{\min }=\mathrm{N}-\mathrm{δ\pm ·} \\ =40-1 \\ =39\mathrm{beats} \end{gathered}$$

The maximum time is,-

$$\begin{gathered} {\mathrm{t}}_{\max }=\mathrm{t}+\mathrm{δ³Ù} \\ =30+0.5 \\ =30.5\sec \end{gathered}$$

The minimum number of beats per minute is define as follows.

$$\begin{gathered} {\mathrm{K}}_{\min }=\frac{{\mathrm{N}}_{\max }}{{\mathrm{t}}_{\min }} \\ =\frac{39\mathrm{beats}}{30.5\sec }×\frac{60\sec }{1\min } \\ =76.8\mathrm{beats}/\min \end{gathered}$$

The uncertainty beats per minute is defined as below.

$$\begin{gathered} {\mathrm{δ°­}}_1={\mathrm{K}}_{\max }-\mathrm{K} \\ =83.4-80 \\ =3.4\mathrm{beats}/\min \\ {\mathrm{δ°­}}_2=\mathrm{K}-{\mathrm{K}}_{\min } \\ =80-76.8 \\ =3.2\mathrm{beats}/\min \end{gathered}$$

The average uncertainty beats per minute is,-

$$\begin{gathered} {\mathrm{δ°­}}_{\mathrm{avg}}=\frac{{\mathrm{δ°­}}_1+{\mathrm{δ°­}}_2}{2} \\ =\frac{3.4+3.2}{2} \\ =3.3\mathrm{beats}/\min \\ ≈3\mathrm{beats}/\min \end{gathered}$$

Therefore, the heart rate is $\mathrm{K}=80\mathrm{beats}/\min$and average uncertainty beats per min is ${\mathrm{δ°­}}_{\mathrm{avg}}=3\mathrm{beats}/\min$.