Question

1. the systolic blood pressure of adults in the usa is nearly normally distributed with a mean of 117 millimeters of mercury (mmhg) and standard deviation of 22.

someone qualifies as having stage 2 high blood pressure if their systolic blood pressure is 160 or higher. stage 1 high bp is specified as systolic bp between 140 and 160. (give answers to four decimal places)
a. what is the probability that an adult in the usa has stage 2 high blood pressure?
b. what is the probability that an adult in the usa has stage 1 high blood pressure?
c. your doctor tells you you are in the 30th percentile for blood pressure among us adults. what is your systolic bp?
d. what is the systolic blood pressure that cuts off the top 2.5% of adults in the usa?

Explanation:

Step1: Calculate z - score for stage 2

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 117\), \(\sigma=22\), and \(x = 160\).

$$z=\frac{160 - 117}{22}=\frac{43}{22}\approx1.9545$$

We want \(P(X\geq160)\), which is \(1 - P(X < 160)\). Using the standard normal table, \(P(Z<1.95)=0.9744\), \(P(Z < 1.96)=0.9750\). Using linear interpolation (or a calculator), \(P(Z<1.9545)\approx0.9746\). So \(P(X\geq160)=1 - 0.9746 = 0.0254\)

Step2: Calculate z - scores for stage 1

For \(x = 140\), \(z_1=\frac{140 - 117}{22}=\frac{23}{22}\approx1.0455\)
For \(x = 160\), \(z_2\approx1.9545\)
\(P(140<X<160)=P(X < 160)-P(X < 140)\)
\(P(Z<1.04)=0.8508\), \(P(Z < 1.05)=0.8531\). Using linear interpolation (or a calculator), \(P(Z<1.0455)\approx0.8520\)
\(P(140<X<160)=0.9746-0.8520 = 0.1226\)

Step3: Find the value for the 30th percentile

If \(P(X < x)=0.30\), we need to find the z - score \(z\) such that \(P(Z < z)=0.30\). Looking up in the standard normal table, \(z\approx - 0.5244\)
Using the formula \(z=\frac{x-\mu}{\sigma}\), we solve for \(x\):
\(-0.5244=\frac{x - 117}{22}\)
\(x=117+(- 0.5244)\times22=117 - 11.5368 = 105.4632\)

Step4: Find the value for the top 2.5%

If \(P(X>x)=0.025\), then \(P(X < x)=0.975\). The z - score corresponding to \(P(Z < z)=0.975\) is \(z = 1.96\)
Using \(z=\frac{x-\mu}{\sigma}\), we solve for \(x\):
\(1.96=\frac{x - 117}{22}\)
\(x=117+1.96\times22=117 + 43.12=160.12\)

Answer:

a. \(0.0254\)
b. \(0.1226\)
c. \(105.4632\)
d. \(160.12\)