Id a data set with unknown shape has a mean 36 and standard deviation 12,then at

Question

Id a data set with unknown shape has a mean 36 and standard deviation 12,then at most percent of observation Would lie outside of the interval ( 12;60) Let's Z~N ( 0;1) p(Z=1) And Z~ uniform (1,5) p( 2.2<x<3.5)
Id a data set with unknown shape has a mean 36 and standard deviation 12,then at most percent of observation Would lie outside of the interval ( 12;60) Let's Z~N ( 0;1) p(Z=1) And Z~ uniform (1,5) p( 2.2<x<3.5)
Id a data set with unknown shape has a mean 36 and standard deviation 12,then at most percent of observation Would lie outside of the interval ( 12;60) Let's Z~N ( 0;1) p(Z=1) And Z~ uniform (1,5) p( 2.2<x<3.5)

Explanation / Answer

(1) Id a data set with unknown shape has a mean 36 and standard deviation 12,then at most percent of observation Would lie outside of the interval ( 12;60)

here we want P(X<12 and X>60)

or, P(X<12 and X>60)=1-P(12<X<60)=1-P(-2<Z<2)

using standard normal variate Z=(X-mean)/sd

for X=12, Z=(12-36)/12=-2

for X=60, Z=(60-36)/12=2

P(-2<Z<2)=P(Z<2)-P(Z<-2)=0.9772-0.0228=0.9544

P(Z<2)=0.9772 ( using ms-excel command =normsdist(2))

-P(Z<-2)=0.0228( using ms-excel command =normsdist(-2))

(second part) Let's Z~N ( 0;1)

p(Z=1)=0 ( since for continuous distribution probability at any point is zero)

if we want P(Z<=1) then it is equalt to 0.8413

(third part) Z~ uniform (1,5)

p( 2.2<x<3.5)=(3.5-2.2)/(5-1)= 0.325

using the uniform distribution property if X~uniform(a,b) then

F(x)=p(X<x)=(X-b)/(b-a) and P(x1<X<x2)=(x2-x1)/(b-a)

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