Asked by Denim
I cant figure this out!!!
The income of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. If a sample of 100 trainees is selected, what is the probability that the sample mean will be less than $1075 month?
The income of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. If a sample of 100 trainees is selected, what is the probability that the sample mean will be less than $1075 month?
Answers
Answered by
PsyDAG
The standard error (SE) of the mean is the standard deviation (SD) divided by the square root of the number in the sample.
In this case, your Z score would be ($1075-$1100)/SE. Find the Z score and look it up in the back of your statistics text in a table labeled something like "areas under the normal disrtibution."
I hope this helps. Thanks for asking.
In this case, your Z score would be ($1075-$1100)/SE. Find the Z score and look it up in the back of your statistics text in a table labeled something like "areas under the normal disrtibution."
I hope this helps. Thanks for asking.
Answered by
Denim
Still don't get it...sorry. I am really trying to get it without asking for the answer, but I just don't get it. Help please.....
Answered by
PsyDAG
Your SE = SD/sq.root of n
SE = 150/10 = 15
For a distribution of sample means, the Z score = (1075-1100)/15.
Can you calculate the Z score and use it in the table to find the probability?
I hope this helps a little more. Thanks for asking.
SE = 150/10 = 15
For a distribution of sample means, the Z score = (1075-1100)/15.
Can you calculate the Z score and use it in the table to find the probability?
I hope this helps a little more. Thanks for asking.
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