) 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 by Guest

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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