The z -score corresponding to an area of 0.99 is 2.33 (invNorm (0.99) on the calculator). So, z (^) x =
2.33. But, also,
Thus,
.
Solving algebraically for x , we get x = 709.7. Rachel needs a score of 710 or higher.
Remember that this type of problem is usually solved by expressing z in two ways (using the
definition and finding the area) and solving the equation formed by equating them.
On the TI-83/84, the answer could be found as follows: invNorm(0.99,500, 90) = 709.37.
(a) μ 3 + 6 X = 3 + 6μ (^) X = 3 + 6(3) = 21.
(b) Because .
Thus,