Question

A certain dataset of systolic blood pressure measurements has a mean of 80 and a standard deviation of 3. Assuming the distribution is bell-shaped and we randomly select a measurement:
a) What percentage of measurements are between 71 and 89?
b) What is the probability a person's blood systolic pressure measures more than 89?
c) What is the probability a person's blood systolic pressure being at most 75?
d) We should expect 15% of patients have a blood pressure below what measurement?
e) Would it be unusual for 3 patients to have a mean blood pressure measurement of more than 84? Explain.

Answer 1

Answer:

Explained below.

Step-by-step explanation:

Let X = systolic blood pressure measurements.

It is provided that, $X\sim N(\mu=80,\sigma^{2}=3^{2})$.

(a)

Compute the percentage of measurements that are between 71 and 89 as follows:

$P(71$

                        $=P(-3$

The percentage is, 0.9973 × 100 = 99.73%.

Thus, the percentage of measurements that are between 71 and 89 is 99.73%.

(b)

Compute the probability that a person's blood systolic pressure measures more than 89 as follows:

$P(X>89)=P(\frac{X-\mu}{\sigma}>\frac{89-80}{3})$

                $=P(Z>3)\\=1-P(Z$

Thus, the probability that a person's blood systolic pressure measures more than 89 is 0.0014.

(c)

Compute the probability that a person's blood systolic pressure being at most 75 as follows:

Apply continuity correction:

$P(X\leq 75)=P(X$

                $=P(X$

Thus, the probability that a person's blood systolic pressure being at most 75 is 0.034.

(d)

Let x be the blood pressure required.

Then,

P (X < x) = 0.15

⇒ P (Z < z) = 0.15

⇒ z = -1.04

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\-1.04=\frac{x-80}{3}\\\\x=80-(1.04\times3)\\\\x=76.88\\\\x\approx 76.9$

Thus, the 15% of patients are expected to have a blood pressure below 76.9.

(e)

A z-score more than 2 or less than -2 are considered as unusual.

Compute the z score for $\bar x$ as follows:

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}$

  $=\frac{84-80}{3/\sqrt{3}}\\\\=2.31$

The z-score for the mean blood pressure measurement of 3 patients is more than 2.

Thus, it would be unusual.