Question

suppose that the heart beat per minute (bpm) of adult males has a normal distribution with a mean of μ = 68 bpm and a standard deviation of σ = 11.5 bpm. instead of using 0.05 for identifying significant values, use the criteria that a value x is significantly high if p(x or greater) ≤ 0.01 and a value is significantly low if p(x or less) ≤ 0.01. 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 □ and less than □ are not significant, and values outside that range are considered significant.
(round to one decimal place as needed.)

Explanation:

Step1: Find the z - score for the lower tail

Since \(P(X\leq x)=0.01\), looking up the z - score in the standard normal table. The z - score \(z_{1}\) corresponding to a left - tail area of \(0.01\) is approximately \(z_{1}=- 2.33\)

We use the z - score formula \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 68\), \(\sigma=11.6\)

Substituting \(z=-2.33\) into the formula: \(-2.33=\frac{x - 68}{11.6}\)

Solving for \(x\): \(x=68+(-2.33)\times11.6=68 - 27.028\approx40.9\)

Step2: Find the z - score for the upper tail

Since \(P(X\geq x)=0.01\), the z - score \(z_{2}\) corresponding to a right - tail area of \(0.01\) (left - tail area \(1 - 0.01=0.99\)) is approximately \(z_{2}=2.33\)

Using the z - score formula \(z=\frac{x-\mu}{\sigma}\), substituting \(z = 2.33\), \(\mu = 68\), \(\sigma=11.6\)

\(2.33=\frac{x - 68}{11.6}\)

Solving for \(x\): \(x=68+2.33\times11.6=68 + 27.028\approx95.0\)

Answer:

A heart rate with a bpm more than \(40.9\) and less than \(95.0\) are not significant.