Question

For women aged 18-24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. If 23 women aged 18-24 are randomly selected, find the probability that their mean systolic blood pressure is between 119 and 122. Your answer should be a decimal rounded to the fourth decimal place.

Answer 1

Answer:

5.77% probability that their mean systolic blood pressure is between 119 and 122.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu - 114.8, \sigma = 13.1, n = 23, s = \frac{13.1}{\sqrt{23}} = 2.73$

Find the probability that their mean systolic blood pressure is between 119 and 122.

This is the pvalue of Z when X = 122 subtracted by the pvalue of Z when X = 119. So

X = 122

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{122 - 114.8}{2.73}$

$Z = 2.64$

$Z = 2.64$ has a pvalue of 0.9959

X = 119

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{119 - 114.8}{2.73}$

$Z = 1.54$

$Z = 1.54$ has a pvalue of 0.9382

So there is a 0.9959 - 0.9382 = 0.0577 = 5.77% probability that their mean systolic blood pressure is between 119 and 122.