SupposeXN(12,32). Find the 90^th percentile ofX . To find your answer, pretend that the following table is (x)=P(Zx), where ZN(0,1). (Note: these values do not agree with the actual normal tables! This artificial table is constructed so that you cannot find the answer in software or elsewhere.)

x	(x)	x	(x)	x	(x)
0.1	0.536	1.1	0.828	2.1	0.934
0.2	0.571	1.2	0.846	2.2	0.937
0.3	0.606	1.3	0.863	2.3	0.940
0.4	0.640	1.4	0.877	2.4	0.943
0.5	0.672	1.5	0.890	2.5	0.944
0.6	0.700	1.6	0.900	2.6	0.946
0.7	0.732	1.7	0.910	2.7	0.947
0.8	0.759	1.8	0.918	2.8	0.948
0.9	0.784	1.9	0.925	2.9	0.948
1.0	0.807	2.0	0.930	3.0	0.950

Group of answer choices

5.6

7.2

8.8

15.8

16.8

18.4

None of the above