Thanks for the supplied normal table; very handy.
We have random variable H with μ=25, σ=6 and we seek P(H < 37).
H=37 corresponds to a z score (number of standard deviations from the mean) of
z = (H - μ)/σ = (37 - 25)/6 = 2
We just found out
P(H < 37) = P(z < 2)
which we can look up. Of course for a nice round z score of 2 we don't really need to look that one up. By the 68-95-99.7 rule, two sigma corresponds to 95%, i.e.
P(-2 < z < 2) = 95%
That means
P(0 < z < 2) = 95/2 = 47.5 %
Of course
P(z < 0) = 50%
so
P(z < 2) = P(z < 0) + P(0 < z < 2) = 50 + 47.5 = 97.5%
Answer: C
For completeness, let's look up z=2 in the table. It doesn't say but this particular table is telling us the answer we want directly.
P(z < 2) = .9772 = 97.72%
which is a bit more accurate than the rule, but in stats that doesn't usually matter much.
