SOLUTION: In a normal distribution with mean 56 and standard deviation 21, how large a sample must be taken so that there will be at least a 90% percent chance that its mean is greater tha
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Question 1192857: In a normal distribution with mean 56 and standard deviation 21, how large a sample must be taken so that there will be at least a 90% percent chance that its mean is greater than
52? Recall that Z= x-mu/sigma/sqrt(n)
Can a tutor please assist me, what should I do to find n? Answer by math_tutor2020(3817) (Show Source):
We're wanting to find the value of n that will make
P(xbar > 52) = 0.90
to be possible.
This is equivalent to finding a value of k such that
P(Z > k) = 0.90
and that in turn is the same as
P(Z < k) = 0.10
I'll use this Z table https://www.ztable.net/
to find that P(Z < -1.28) = 0.10027 which is the closest we can get to 0.10
So P(Z < -1.28) = 0.10 approximately
and
P(Z > -1.28) = 0.90 approximately
The key takeaway is that z = -1.28 is the approximate z score.
That z score is then used to solve for n like so
Z = (x-mu)/(sigma/sqrt(n))
-1.28 = (52-56)/(21/sqrt(n))
-1.28*(21/sqrt(n)) = -4
-26.88 = -4sqrt(n)
sqrt(n) = -26.88/(-4)
sqrt(n) = 6.72
n = (6.72)^2
n = 45.1584
n = 45
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