The heights of adult women are approximately normally distributed about a mean of 65 inches with a standard deviation of 2 inches. If Rachael is at the 99th percentile in height for adult women, then her height, in inches, is closest to

a) 60
b) 62
c) 68
d) 70
e) 74

I know that the 99th percentile means that she is better than 99 percent of the population. But I'm not sure where to go or think from there.

Question

The heights of adult women are approximately normally distributed about a mean of 65 inches with a standard deviation of 2 inches.  If Rachael is at the 99th percentile in height for adult women, then her height, in inches, is closest to
 
a) 60
b) 62
c) 68
d) 70
e) 74
 
I know that the 99th percentile means that she is better than 99 percent of the population. But I'm not sure where to go or think from there.

displayerror · Answer

You have to make use of the z-table (a copy of it is here https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf ). For the 99th percentile, we look on the table for the value closest to 0.99. That z-value corresponds to the 99th percentile (I got something around 2.33, but that isn't the exact value). We can then use that z-value in the formula provided here http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_Probability/BS704_Probability10.html to solve for the height that is at the 99th percentile by plugging in all of the given information.