The mean per capita income is 16,44516,445 dollars per annum with a standard deviation of 397397 dollars per annum. What is the probability that the sample mean would differ from the true mean by greater than 3838 dollars if a sample of 208208 persons is randomly selected? Round your answer to four decimal places.

Hence the probability of the getting Sample mean and true mean differ greater than by  38 dollars is 0.91621

Step-by-step explanation:

Given:

(The values of mean and the difference in sample mean and true mean is 3838 which not accountable with respect to highly numbered  mean

and 0.0001266 standard deviation which is not  affected to that value in terms of Z-score and P value)

Hence consider

Mean as =16445 dollars,

Standard deviation=397(2.41 %  which seems valid)

No of persons:208

To  Find:

And sample mean and true mean differ by 38 or more

i.e (sample mean -true mean )≥38.

Solution:

So calculate Z value as,

Z=(-38)/(397/Sqrt(208))

=-(38*14.4222)/397

=-1.38042

Now it should be Greater than this value,

Using Z-table we get P value  as follows

Hence P(Z≥-1.38042)

=0.91621

Hence the probability of the getting Sample mean and true mean differ greater than by  38 dollars is 0.91621

the answer to this question is yes

step- i d k and i d c : )

you: : 0

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