# Solved (Free): Our subjects are males aged 35-44 whose blood pressures are normally distributed with mean 80 mmHg and standard deviation 12

#### ByDr. Raju Chaudhari

Apr 3, 2021

Our subjects are males aged 35-44 whose blood pressures are normally distributed with mean 80 mmHg and standard deviation 12 mmHg.

a) Someone who is borderline hypertensive has a diastolic blood pressure between 90 and 95 mmHg inclusive; what proportion of these subjects are borderline hypertensive?
b) Someone who is hypertensive has a diastolic blood pressure above 95 mmHg; what proportion of these subjects are hypertensive?

#### Solution

Given that $\mu = 80$ and $\sigma = 12$.

a) Someone who is borderline hypertensive has a diastolic blood pressure between 90 and 95 mmHg inclusive; the proportion of these subjects are borderline hypertensive

 \begin{aligned} P(90 < X < 95) &= P(X < 95) - P(X < 90)\\ &= P\bigg(\frac{X-\mu}{\sigma} < \frac{95-80}{12}\bigg)-P\bigg(\frac{X-\mu}{\sigma} < \frac{90-80}{12}\bigg)\\ &=P\bigg(Z < 1.25\bigg)-P\bigg(Z < 0.83\bigg)\\ &=0.8944-0.7967\\ &= 0.0976 \end{aligned}

b) Someone who is hypertensive has a diastolic blood pressure above 95 mmHg; the proportion of these subjects are hypertensive is

 \begin{aligned} P(X > 95) &= 1-P(X\leq95)\\ &= 1- P\bigg(\frac{X-\mu}{\sigma} \leq \frac{95-80}{12}\bigg)\\ &=1- P\bigg(Z \leq 1.25\bigg)\\ &=1-0.8944\\ &= 0.1056 \end{aligned}