SOLUTION: The mean rent of a 3-bedroom apartment in Orlando is $1200. You randomly select 12 apartments around town. The rents are normally distributed with a standard deviation of $300. Wha

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Question 1104234: The mean rent of a 3-bedroom apartment in Orlando is $1200. You randomly select 12 apartments around town. The rents are normally distributed with a standard deviation of $300. What is the probability that the mean rent is more than $1100?
0.1827
0.8173
0.1251
0.8749
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
mean = 1200
standard deviation = 300
sample size = 12

standard error = standard deviation of population / square root of sample size = 300 / sqrt(12) = 86.60254038.

z-score = (x-m)/s

x is the raw score
m is the raw mean
s is the standard error.

z = (1100 - 1200) / (300/sqrt(12)) = -1.154700538

using a z-score calculator, the probability of getting less than a z-score of -1.154700538 is equal to .1241065934.

the probability of getting more than a z-score of -1.154700538 is equal to 1 - .124065934 = .8758934066.

the closest selection is .8749.

your problem more then likely used rounded numbers so i would guess that selection 4 is correct.

if i rounded the z-score to -1.15, i would get .8749280114 which equals .8749 rounded to 4 decimal places.

confirmation that this is correct can be made through this online calculator at http://davidmlane.com/hyperstat/z_table.html

first picture is the raw score results.
second picture is the not rounded z-score results.
third picture is the rounded z-score results.

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as you can see, raw results and no rounding of the z-score results in .8759, while rounding the z-score results in .8749.

if you solve the problem and you're close to one of the results, but not right on, consider the difference may possibly be due to rounding.