Using a normal curve table, if a person has a music aptitude score of 41, which equals a z score of 1.3, the percentage of people having a higher score is

To answer this problem, we simply have to refer to the standard normal probabilities table to locate for the P value at specified z score value.

So at a value of z = 1.3, the value of P using right tailed test is:

P = 0.0968 = 9.68%

So this actually means that 9.68% of the people has a higher score

452,489 to the nearest thousand is 452,000