Answer:
(a) 81st
=NORM.INV(0.81;0;1)
[tex] 0.878[/tex]
(b) 19th
=NORM.INV(0.19;0;1)
[tex] -0.878[/tex]
(c) 76th
=NORM.INV(0.76;0;1)
[tex] 0.706[/tex]
(d) 24th
=NORM.INV(0.24;0;1)
[tex] -0.706[/tex]
(e) 11th
=NORM.INV(0.11;0;1)
[tex]-1.227[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
For this case we can find the percentiles with the Norm.inv function in Excel and we got:
(a) 81st
=NORM.INV(0.81;0;1)
[tex] 0.878[/tex]
(b) 19th
=NORM.INV(0.19;0;1)
[tex] -0.878[/tex]
(c) 76th
=NORM.INV(0.76;0;1)
[tex] 0.706[/tex]
(d) 24th
=NORM.INV(0.24;0;1)
[tex] -0.706[/tex]
(e) 11th
=NORM.INV(0.11;0;1)
[tex]-1.227[/tex]
