Question 324439: Z Test question....
A state community is elgible for clothes the mean income per annum is $3,000 or less. A simple random sample of 100 people was drawn from the place and the mean income is $2,500. The standard deviation is $200. Find out the proportion with annual incom between $2500 or less?
This is what I got... 3000-2500/200= 500/200= 2.5(Check the appendix for z at 2.5) I got .4938 and .5-.4938=.0062 when converted to percent I got .62%
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! Your question is confusing.
Find out the propoetion with annual income between $2500 and less.
Did you mean $2500 and less?
If so why did you use $3000 to calculate the z score?
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Also, if z=2.5, then P=0.5+0.4938=0.9938, add, don't subtract.
So if the mean is $2500 and std. dev. is $200, 99.38% of the people make less than $3000.
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Please re-post.
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