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October 19th, 2016, 02:48 PM   #1
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Probability & Central Limit Theorem

The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual.
μx̄ = μ = 12,749
σ = 1.2
n = 35

For the given sample n = 35, the probability of a sample mean being less than 12,749 or greater than 12,752 is ____________


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October 20th, 2016, 04:21 PM   #2
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This is what I have so far.
σx = σ/ √n = 1.2/ √35 = 0.2028

z = x̄ - μx̄/σx = x̄ - μx̄/σ/√n = 12,749 - 12,752/ 1.2√35 = -0.4226 = .33724

I stopped right there because I got confused. I'm stuck.
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October 21st, 2016, 01:10 AM   #3
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Quote:
Originally Posted by rihnavy View Post
This is what I have so far.
σx = σ/ √n = 1.2/ √35 = 0.2028

z = x̄ - μx̄/σx = x̄ - μx̄/σ/√n = 12,749 - 12,752/ 1.2√35 = -0.4226 = .33724

I stopped right there because I got confused. I'm stuck.
The Z-value should be -14.79, not -.4226 - you might want to check that you did not erroneously punch the '*' button instead of '/'.
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October 23rd, 2016, 07:57 AM   #4
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Is the first equation that I have correct or should it be z= 12749 - 12749/1.2/√35 = 0
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October 23rd, 2016, 02:28 PM   #5
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Quote:
Originally Posted by rihnavy View Post
This is what I have so far.
σx = σ/ √n = 1.2/ √35 = 0.2028

z = x̄ - μx̄/σx = x̄ - μx̄/σ/√n = 12,749 - 12,752/ 1.2√35 = -0.4226 = .33724
You're missing grouping symbols.

z = (x̄ - μx̄)/(σ/√n)

According to the problem, μx̄ = 12,749, not 12,752.


If x̄ is supposed to equal 12,752, then

z = (12,752 - 12,749)/(1.2/√35) ~ 14.79


That would be one part of the problem.
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