Question

assume that blood pressure readings are normally distributed with a mean of 122 and a standard deviation of 4.8. if 30 people are randomly selected, find the probability that their mean blood pressure will be less than 124, rounding to four decimal places.

a. 0.9908
b. 0.0062
c. 0.0615
d. 0.9938

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 4.8$ and $n = 30$, we have $\sigma_{\bar{x}}=\frac{4.8}{\sqrt{30}}\approx0.876$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x}=124$, $\mu = 122$, and $\sigma_{\bar{x}}\approx0.876$. So $z=\frac{124 - 122}{0.876}\approx2.28$.

Step3: Find the probability

We want $P(\bar{X}<124)$. Using the standard normal distribution table (or a calculator with a normal - distribution function), $P(Z < 2.28)$. From the standard normal table, $P(Z<2.28)=0.9887 + 0.0051=0.9938$ (using the fact that for $z = 2.28$, we can also use a calculator: $P(Z<2.28)=\text{normalcdf}(-\infty,2.28,0,1)\approx0.9887 + 0.0051 = 0.9938$).

Answer:

D. 0.9938