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mean = 75; variance = 18.49, if a person is selected at random, what is the probability that their resting pulse rate is under 70?
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Assuming a normal distribution:\[\sigma=\sqrt{18.49}=4.3\]\[Z=\frac{70-75}{4.3}\approx-1.16\]\[P(Z<-1.16)=1-P(Z<1.16)=1-\Phi(1.16)=1-0.877=0.123\]
why can't you just use normal cdf (0,70,75,18.490?
You can, that's what I did. Maybe you're not familiar with the notation?
oh you know what .. i assumed that 18.49 was the standard deviation but you square rooted it... why do you square root it for? is it because it is the "variance?"
Yes, variance is defined as standard deviation squared.
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ok thanks for clarifying that for me :) i get it now
Glad to hear it.
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