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the lengths of individual fish in a pond have an approximate normal distribution with a mean of 37.2 cm and a standard deviation of 12.0 cm. a sample of 36 fish is taken. what is the standard deviation of the sampling distribution of the sample mean for the 36 fish?



Answer :

ANKITANAND

The standard deviation of the sampling distribution of the sample mean for the 36 fish will be less than 0.0003.

What is standard deviation?

The square root of the variance is used to calculate the standard deviation, a statistic that expresses how widely distributed a dataset is in relation to its mean.

standard deviation of sample mean = sigma X'=sigma/√n=2

according to the question,

we are given the mean to be 37.2

standard deviation=12.0

sample=36

n=(6/0.5)2 =144

n=p(1-p)/(0.05)2 =0.2*0.8/(0.05)2 =64

for normal distribution z score =(p'-p)/sigma p      

       

here population proportion=     p= 0.510      

       

sample size       =n= 600      

std error of proportion=σp=√(p*(1-p)/n)= 0.0204      

       

probability = P(X<0.42) = P(Z<-4.41)= 0.0000

therefore less than 0.0003 is correct.

Therefore, the standard deviation of the sampling distribution of the sample mean for the 36 fish will be less than 0.0003

to learn more about standard deviation visit:

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