According to a study, the scale of scores on an iq test of adults is approximately normal with mean 98 and standard deviation 13 . the organization mensa, which calls itself "the high-iq society," requires an iq score of 130 or higher for membership. what percent of adults (± 0.1%) would qualify for membership?

First you need to find your z score of 130 on this normal curve. The formula for z-score is (x-mean)/standard deviation.
So in this case: z=(130-98)/13=2.461
Using your zchart for a normal curve we find that the area under the curve for this zscore is .9931. 
However, since we are looking for 130 or higher, we must subtract this value from 1
So 1-.9931=.0069
This is the area to the right of the curve and represents the percentage of people with an IQ greater than or equal to 130
The answer should be .69% (+/- 1%)