Question

suppose that the heart - beat per minute (bpm) of adult males has a normal distribution with a mean of $mu = 70.4$ bpm and a standard deviation of $sigma = 11.1$ bpm. instead of using 0.05 for identifying significant values, use the criteria that a value $x$ is significantly high if $p(x greater)leq0.01$ and a value is significantly low if $p(x or less)leq0.01$. find the pulse rates for males that separate significant pulse rates from those that are not significant. using these criteria, is a male pulse rate of 90 beats per minute significantly high?
find the heart rate (in bpm) separating significant values from those that are not significant.
a heart rate with a bpm more than $square$ and less than $square$ are not significant, and values outside that range are considered significant.
(round to one decimal place as needed.)

Explanation:

Step1: Find z - scores for significant values

We use the standard normal distribution table. For a left - tailed probability of 0.01, the z - score $z_1$ is approximately - 2.33. For a right - tailed probability of 0.01, the z - score $z_2$ is approximately 2.33.

Step2: Use z - score formula to find heart rates

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the original distribution, $\mu$ is the mean, and $\sigma$ is the standard deviation. We want to solve for $x$. Rearranging the formula gives $x=\mu + z\sigma$.
For the lower significant value:
$x_1=\mu+z_1\sigma$, substituting $\mu = 70.4$, $z_1=-2.33$, and $\sigma = 11.1$.
$x_1=70.4+( - 2.33)\times11.1=70.4 - 25.863=44.537\approx44.5$.
For the upper significant value:
$x_2=\mu+z_2\sigma$, substituting $\mu = 70.4$, $z_2 = 2.33$, and $\sigma = 11.1$.
$x_2=70.4+2.33\times11.1=70.4 + 25.863=96.263\approx96.3$.

Answer:

A heart rate with a bpm more than 44.5 and less than 96.3 are not significant, and values outside that range are considered significant.