Respuesta :
The interpolations;
(a) 81st
= NORM.INV(0.81;0;1)
0.878
(b) 19th
=NORM.INV(0.19;0;1)
-0.878
(c) 76th
=NORM.INV(0.76;0;1)
0.706
(d) 24th
=NORM.INV(0.24;0;1)
-0.706
(e) 11th
=NORM.INV(0.11;0;1)
-1.227
Given,
Data near the mean are more likely to occur than data distant from the mean, according to the normal distribution, which is a probability distribution that is symmetric around the mean.
"A numerical measurement used in statistics of a value's relationship to the mean (average) of a set of values, assessed in terms of standard deviations from the mean," according to Wikipedia, is the Z-score.
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)
0.878
(b) 19th
=NORM.INV(0.19;0;1)
-0.878
(c) 76th
=NORM.INV(0.76;0;1)
0.706
(d) 24th
=NORM.INV(0.24;0;1)
-0.706
(e) 11th
=NORM.INV(0.11;0;1)
-1.227
Learn more about interpolations here;
https://brainly.com/question/15113672
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