SOLUTION: A normal population has a mean of $67 and standard deviation of $19. You select random samples of nine.
a.What is the standard error of the sampling distribution of sample means
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Question 1192814: A normal population has a mean of $67 and standard deviation of $19. You select random samples of nine.
a.What is the standard error of the sampling distribution of sample means?
b.What is the probability that a sample mean is greater than $70?
c.What is the probability that a sample mean is less than $63?
d. What is the probability that a sample mean is between $63 and $70?
e.What is the probability that the sampling error ( x⎯⎯ − μ) would be $9 or more? That is, what is the probability that the estimate of the population mean is less than $58 or more than $76?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
It is $19/sqrt(9), the sd/sqrt(n)=$6.33
b. z=(x-mean)/sigma/sqrt(n)=3/(19/3)=0.4739
prob. z >0.4739=0.3178
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c. This is z <-4/6.33 =-0.6319; probability of that is 0.2637
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d.Between 63 and 70 is 1- the sum of those two probabilities above, or 1-0.5815=0.4185
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e. This would be +/- 9/6.33=1.42. That probability is 1-0.8444 (the probability between them)=0.1556.
There will be slight differences in the fourth decimal place if you round the z-value. Generally, 2 places may be done, but some may want you to round the z-value only at the end.
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