# ) 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. a) If a woman between the ages of 18 and 24 is randomly selected, find the probability that her systolic blood pressure is greater than 125.

**Answer:**

**0.2177**

**Explanation:**

The probability distribution of the standard normal variable, Z, is tabulated.

Z, the standard normal variable, is defined by:

**Z = (X - μ) / σ**, where

- X is a normal variable (the systolic blood pressure,in mm Hg in this case)

- μ is the mean (114.8 mm Hg in this case), and

- σ is the standard deviation (13.1 mm Hg in this case).

You want to *find the probablity that the systolic pressure of a woman between the ages of 18 and 24 is greater than 125*, which means **P (X > 125).**

Then, to use a **table of Z-score**, you have to convert X > 125 into Z and find the corresponding probabiiity.

These are the calculations:

- X > 125 ⇒ Z > (125 - 114.8) / 13.1 ⇒ Z > 10.2 / 13.1 ⇒ Z > 0.7786

Now use a table for the normal standard probabiity. Most tables use two decimals for Z, so you can round to Z > 0.78, which will yield P (Z > 0.78) = 0.2177.

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