SOLUTION: Given that you have a mean of 100 and standard deviation of 3; if repeated samples of 36 were selected what proportion of them would have means less than 99 and greater than 100?

Question 268464: Given that you have a mean of 100 and standard deviation of 3; if repeated samples of 36 were selected what proportion of them would have means less than 99 and greater than 100?
I have tried to solve this using central limit therom and end up with
-1/(3/6) = -1/0.5 = -0.5 however the book states that
SE with n=36 is 3/6 The critical ratio for a mean equal to 99 is (99-100)/0.5=-2.00 and for 101 is +2.00
I do not understand the critical ration portion...if you solve the equation the answer is -0.5 not -2.00...where am I going wrong?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Given that you have a mean of 100 and standard deviation of 3; if repeated samples of 36 were selected what proportion of them would have means less than 99 and greater than 101?
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t(99) = (99-100/(3/sqrt(36)) = -1/(1/2) = -2
t(101) = (101-100)/(3/sqrt(36)) = 1/(1/2) = +2
================
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P(xbar< 99) = P(t< -2 when df = 35) = 0.0267
P(xbar> 101) = P(t > 2 when df = 35) = 0.0267
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Comment: You just made an arithmetic error. Statistics
is full of these traps.
==================================================
Cheers,
Stan H.

Given that you have a mean of 100 and standard deviation of 3; if repeated samples of 36 were selected what proportion of them would have means less than 99 and greater than 100?
I have tried to solve this using central limit therom and end up with
-1/(3/6) = -1/0.5 = (-0.5 WRONG) however the book states that
SE with n=36 is 3/6 The critical ratio for a mean equal to 99 is (99-100)/0.5=-2.00 and for 101 is +2.00 (CORRECT)
I do not understand the critical ration portion...if you solve the equation the answer is -0.5 not -2.00...where am I going wrong?

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