Question

Suppose you are taking a random sample, and measuring continuous variable X, from a population in which X is normally distributed. What is the probability that you will get a sample mean greater than the 60th percentile of the original distribution of individuals,
i. with a sample size of N=100?
ii. with a sample size N=50?
iii. with a sample size of N=10?
Do any of your answers change if the original distribution is not normal? Explain.

Answer 1

i) for z score =(X-mean)/std deviation

as we know that for 60th percentile z score=0.25

hence (Xbar-) =0.25*

for sample size n=100 ; std error =/sqrt(n) =/sqrt(100) =0.1

hence P( sample mean greater than the 60th percentile of the original distribution of individuals)

=P(Xbar- >0.25*)=P(Z>0.25*/(0.1))=P(Z>2.5)=0.0062

ii)

as above :

for sample size n=50 ; std error =/sqrt(n) =/sqrt(50) =0.1

P(Xbar- >0.25*)=P(Z>0.25*/(0.1414))=P(Z>1.77)=0.0384

iii)

for sample size n=50 ; std error =/sqrt(n) =/sqrt(10) =0.316

P(Xbar- >0.25*)=P(Z>0.25*/(0.316))=P(Z>0.79)=0.2148

as we have given that population is normally distributed; therefore sampling distribution of mean also should be normal irrespective of sample size

but if population is not normal we can not assume in part iii) that sampling distribution of mean is normal as sample size is less than 30 ; also 60th percentile of original population would be different too

therefore Yes our answers change if the original distribution is not normal