Question

assume that blood pressure readings are normally distributed with a mean of 118 and a standard deviation of 4.8. if 36 people are randomly selected, find the probability that their mean blood pressure will be less than 120.

a. 0.8819
b. 0.9938
c. 0.0062
d. 0.8615

Explanation:

Step1: Calculate the standard error

The formula for standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \), where \( \sigma = 4.8 \) and \( n = 36 \).
\( \sigma_{\bar{x}}=\frac{4.8}{\sqrt{36}}=\frac{4.8}{6} = 0.8 \)

Step2: Calculate the z - score

The formula for z - score is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \), where \( \bar{x} = 120 \), \( \mu=118 \) and \( \sigma_{\bar{x}} = 0.8 \)
\( z=\frac{120 - 118}{0.8}=\frac{2}{0.8}=2.5 \)

Step3: Find the probability

We want to find \( P(\bar{X}<120) \), which is equivalent to \( P(Z < 2.5) \)
Looking up the value in the standard normal table, \( P(Z < 2.5)=0.9938 \)

Answer:

B. 0.9938