Question

A blood pressure test was given to 450 women ages 20 to 36. It showed that their mean

systolic blood pressure was 119.4 mm Hg, with a standard deviation of 13.2 mm Hg.

a. Determine the z-score, to the nearest hundredth, for a woman who had a systolic blood

pressure reading of 110.5 mm Hg.

b. The z - score for one woman was 2.15. What was her systolic blood pressure reading?

Answer 1

(a)

Consider the values

$\begin{aligned} &\bar{x}=119.4, \\ &x=110.5, \\ &s=13.2 \end{aligned}$

Substitute the given values into z-score equation and solve it.

$\begin{aligned} Z_{x} &=\frac{x-\bar{x}}{s} \\ &=\frac{110.5-119.4}{13.2} \\ &=-0.67 \end{aligned}$

The z-score is -0.67.

(b)

Consider the values

$\begin{aligned} &z_{x}=2.15 \\ &\bar{x}=119.4 \\ &s=13.2 \end{aligned}$

Substitute the given values into z-score equation and solve for x.

$\begin{aligned} Z_{x} &=\frac{x-\bar{x}}{s} \\ 2.15 &=\frac{x-119.4}{13.2} \\ 28.38 &=x-119.4 \\ 147.78 &=x \end{aligned}$

The systolic blood pressure reading is $147.78 \mathrm{~mm} \mathrm \ {Hg}$