SOLUTION: If the mean and standard deviation of serum iron values for healthy men are 116 and 14 micrograms per 100 mL, respectively, find the probability that a random sample of 58 healthy

Algebra ->  Probability-and-statistics -> SOLUTION: If the mean and standard deviation of serum iron values for healthy men are 116 and 14 micrograms per 100 mL, respectively, find the probability that a random sample of 58 healthy       Log On


   

Question 358105: If the mean and standard deviation of serum iron values for healthy men are 116 and 14 micrograms per 100 mL, respectively, find the probability that a random sample of 58 healthy men will yield a mean of at most 120 micrograms per 100 mL.

I know that N = 58...I think...and I don't know if this makes sense but is 14/sqrt(58) on the right track?
If you could please help, that would be extremely appreciated, I am just completely lost! THANK YOU SO MUCH!
Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
you are given the following
The population mean mu=116 and population standard deviation sigma=14, the population distribution is not known
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also you sample n=58 healthy men
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You want to know
P(xbar<=120)
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you are left with a dilemma. You have partial information, mu and sigma but no knowledge of the parent population. What do you do???
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Well you rely on the Central Limit Theorem that says:
If you are analzying sample averages "and" the sample sizes used for those averages is large (ie over 30), the distribution of the sample averages will resemble a normal curve, specially as the sample sizes get larger.
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Since n=56 can be consider large and you are analzying sample averages you get a free pass to use the normal distribution for your problem
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Your next step would be to standardize the sample averages you are interested in, xbar<=120. Standardize implies you change the values so that they mimick coming from a distribution having mu=0 instead of mu=116 and sigma=1 instead of 14. How do you do that?
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To standarize you use the transformation Z=%28xbar+-mu%29%2Fstderror, where the standard error = sigma%2Fsqrt%28n%29. you have to use standard error because you are analysing sample averages xbar, not individual valus x.
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So your problem P(Xbar<=120) becomes P%28Z%3C=%28120-116%29%2F%2814%2Fsqrt%2858%29%29%29 or P(Z<=2.176)
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from here you can either use book tables, statistical calculators or excel among many other tools to find this probability.
answer=0.9852