The resting heart rate of us females 20 years old or older is normally distributed with a mean of 75 bpm (beats per minute). suppose the standard deviation was 15 bpm. if one randomly selected female's resting heart rate falls in the bottom 33% of all women, at most what could her rate have been?
a. 70.05
b. 68.4
c. 79.95
d. 84.439
e. 6
f. 65.561

69 bpm

Step-by-step explanation:

Here we start out finding the z-score corresponding to the bottom 33% of the area under the standard normal curve.  Using the invNorm( function on a basic TI-83 Plus calculator, I found that the z-score associated with the upper end of the bottom 33% is -0.43073.

Next we use the formula for z score to determine the x value representing this woman's heart rate:

       x - mean                              x - 75 bpm

z = = -0.43073 =

       std. dev.                                    15

Thus,  x - 75 = -0.43073(15) = -6.461, so x = 75 - 6.6461, or approx. 68.54, or (to the nearest integer), approx 69 bpm

you can't solve x because it is 2 linear equation and you only have one equation

but which class are u may know that first