Question 1185696: 2.
A blood pressure test was given to 450 women ages 20 to 36. It showed that their mean systolic blood pressure was 119.3 mm Hg, with a standard deviation of 13.6 mm Hg.
a. Determine the z-score, to the nearest hundredth, for a woman who had a systolic blood pressure reading of 112.8 mm Hg.
b. The z-score for one woman was 2.15. What was her systolic blood pressure reading? (Round your answer to one decimal place.)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**a. Finding the z-score:**
The z-score formula is:
z = (x - μ) / σ
Where:
* x is the individual's blood pressure (112.8 mm Hg)
* μ is the population mean blood pressure (119.3 mm Hg)
* σ is the population standard deviation (13.6 mm Hg)
Plugging in the values:
z = (112.8 - 119.3) / 13.6
z = -6.5 / 13.6
z ≈ -0.48
So, the z-score for a woman with a blood pressure of 112.8 mm Hg is approximately **-0.48**.
**b. Finding the blood pressure:**
We can rearrange the z-score formula to solve for x (the blood pressure):
x = μ + zσ
We're given:
* z = 2.15
* μ = 119.3 mm Hg
* σ = 13.6 mm Hg
Plugging in the values:
x = 119.3 + (2.15 * 13.6)
x = 119.3 + 29.24
x ≈ 148.5
So, the woman with a z-score of 2.15 had a systolic blood pressure reading of approximately **148.5 mm Hg**.
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