For women aged $18 - 24$, systolic blood pressures (in $\mathrm { mm } \mathrm { Hg }$ ) are normally distributed with a mean of $114.8$ and a standard deviation of 13.1. If 36 women are selected at random from a population of 300 women aged 18-24, find the probability that their mean systolic blood pressure will be less than $110 \mathrm {~mm} \mathrm { Hg }$. Assume that the sampling is done without replacement and use a finite population correction factor with $\mathrm { N } = 300$.

Question

Quizplus · Accepted Answer

To find the probability that the sample mean is less than 110 mm Hg, we first calculate the standard error of the mean using the finite population correction factor. The standard error is given by $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}}$, where $\sigma = 13.1$, $n = 36$, and $N = 300$. This gives $\sigma_{\bar{x}} \approx 2.06$. The z-score for 110 is $\frac{110 - 114.8}{2.06} \approx -2.33$. Using the standard normal distribution table, the probability corresponding to a z-score of -2.33 is approximately 0.0096.

For Women Aged 18−2418 - 2418−24 , Systolic Blood Pressures (In mmHg\mathrm { mm } \mathrm { Hg }mmHg

For Women Aged $18 - 24$ , Systolic Blood Pressures (In $\mathrm { mm } \mathrm { Hg }$