1 Answer

Suchith · Answer 1 · 2023-07-01T08:02:43+0000

Let X be the number of heartbeats of any individual in the group.

The variable X is normally distributed with mean 74 beats per minute,

$\mu=74$

The standard deviation is 2 beats per minute,

$\sigma=2$

Consider the formula of z-score corresponding to any value of X=x as,

$z=(x-\mu)/(\sigma)$

It is required to find the probability that the number of heartbeats of a randomly selected person lies between 71 and 75.

This can be obtained as follows,

$\begin{gathered} P(71From the Standard Normal Distribution Table,[tex]\begin{gathered} \phi(0.5)=0.1915 \\ \phi(1.5)=0.4332 \end{gathered}$

Substitute the values and simplify,

$\begin{gathered} =0.4332+0.1915 \\ =0.6247 \end{gathered}$

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