we need to determine P (x> 73)
when
mean: μ = 74 beats/min
standard deviation: σ = 2 beats/min
First we need to use the following formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]where
x = 73
μ = 74
σ = 2
and
Z is the z-score
... therefore
[tex]z=\frac{73-74}{2}=-\frac{1}{2}=-0.5[/tex]If we check a table of z scores, we will find that when z = -0.5, then P = 0.3085
Now, since we need P(x>73)
therefore
[tex]P=1-0.3085=0.6915[/tex]P(x>73) = 0.6915
