The resting heart rate, H beats per minute, of an adult patient at a clinic is modelled by H N(72, 5^2). What is the probability that a randomly chosen patient has a resting heart rate of less than 78 beats per minute?

Answer: A. Standardising H N(72,5^2), so =72 and =5. Standardise using Z=H- /: z=78-72/5=6/5=1.2 Reading the table P(H 78), the…

The Normal Distribution: Question 1

Syllabus 5.5

Multiple choice AS 1 mark

The resting heart rate, $H$ beats per minute, of an adult patient at a clinic is modelled by $H\sim N(72, 5^2)$ .

What is the probability that a randomly chosen patient has a resting heart rate of less than $78$ beats per minute?

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$H\sim N(72,5^2)$ , so $\mu=72$ and $\sigma=5$ . Standardise using $Z=\dfrac{H-\mu}{\sigma}$ :

$z=\frac{78-72}{5}=\frac{6}{5}=1.2$

$P(H<78)=P(Z<1.2)=\Phi(1.2)$

From the standard normal table, $\Phi(1.2)=0.8849$ .

$0.1151=1-\Phi(1.2)$ is $P(H>78)$ , the complement of what is asked.
$0.5000=\Phi(0)$ would only be correct if $78$ bpm were the mean.
$0.9918=\Phi(2.4)$ comes from using $z=2.4$ , e.g. from an incorrectly halved standard deviation of $2.5$ .

$P(H<78)=\boxed{0.8849}$ , so the correct option is A.