Answer to Question #180930 in Statistics and Probability for Richard Sasunathan

Question

Question 2

In a study of hypertension and optimal treatment conducted by the National Heart Institute, 10,000 patients had a mean systolic blood pressure (BP), 𝝁 = 𝟏𝟔𝟏 mm Hg and standard deviation, 𝝈 = 𝟐𝟓 mm Hg. Assume the systolic blood pressure is normally distributed.

a. What is the probability of patients with a systolic blood pressure of more than 180 mm Hg?

b. How many patients will have a systolic blood pressure of more than 180 mm Hg?

c. What is the probability of patients with a systolic blood pressure between 145 and 160 mm Hg?

d. If 60 random samples each of size 30 are drawn from this population, determine:

  i. the sampling distribution of the mean systolic blood pressure.

  ii. the probability that the of mean systolic blood pressure between 140 and 165 mm Hg.

Answer 1

Let X denotes the systolic blood pressure. The X follows normal with $\mu = 161, \sigma = 25$

Let $Z = \frac{X - \mu}{\sigma}$ ~N(0,1)

a. P(X>180)

$P(X>x) = P(\frac{X-\mu}{\sigma}>\frac{x-\mu}{\sigma}) = P(Z>\frac{x-\mu}{\sigma}) = 1 - P(Z\le \frac{x-\mu}{\sigma})$

Therefore, when x = 180,

$\frac{x-\mu}{\sigma}=\frac{180-161}{25}=0.76 \\ \therefore P(X>180) = 1 - P(Z\le 0.76) = 1 - 0.7764 = 0.2236$

b. Number of patients will have systolic blood pressure more than $180 = 0.2236 \times 10000 = 2236$

c. $P(145 \le X \le 160)$

Now,

$P(x_1\le X \le x_2) = P(\frac{x_1-\mu}{\sigma}\le \frac{X-\mu}{\sigma} \le \frac{x_2-\mu}{\sigma}) \\ P(x_1\le X \le x_2) = P(\frac{x_1-\mu}{\sigma}\le Z \le \frac{x_2-\mu}{\sigma}) \\ P(x_1\le X \le x_2) = P(Z \le \frac{x_2-\mu}{\sigma}) - P(Z \le \frac{x_1-\mu}{\sigma})$

when x = 145,

$\frac{x-\mu}{\sigma}=\frac{145-161}{25}=-0.64$

when x = 160,

$\frac{x-\mu}{\sigma}=\frac{160-161}{25}=-0.04$

$P(145 \le X \le 160) = P(Z\le-0.04) - P(Z\le -0.64) = 0.2229$

d.

i. The sampling distribution for mean with sample size $= 60 \times 30 = 1800$ is

$N(161,\sqrt{1800}) = N(161,42.4264)$

ii. $P(X\le 165)$

$P(X\le x) = P(\frac{X-\mu}{\sigma}\le \frac{x-\mu}{\sigma}) = P(Z\le \frac{x-\mu}{\sigma})$

When x = 165

$\frac{x-\mu}{\sigma}=\frac{x-161}{42.4264}=0.4478 \\ \therefore P(Z\le 0.4478) = 0.5376$