Statistics Help please!! +K

coop-son

coop-son

Member
Let μ be the mean annual salary of Major League Baseball players for 2009. Assume that the standard deviation of the salaries of these players is $2,845,000. What is the probability that the 2009 mean salary of a random sample of 26 baseball players was within $500,000 of the population mean, μ? Assume that n/N ≤ .05.

Answer:

.6318

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How did they get that answer?? Please and thank you

+K
 
you could be tricking me...which would be clever, but if you're for real, please type it again! Im desperate

HWW....its okay man
 
I thought Id have to use the Z-score table at some point, but what am I supposed to do with the numbers to GET to the point where i Use the Z table
 
There's a formula for this. On my phone so I can't help. But test the one sided hypothesis with the foothills you're given. This is simple
 
How I know OP has no friends:

He comes to NS for homework help instead of calling someone in his class who probably already has it done and could easily explain it.

Nice.
 
I don't know of this has already been answered but from what I think if it's a normal distribution then you find the two z scores and you subract the larger one from the smaller one hope this helps
 
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that is your formula.

The population mean (mu) is not given, but that isn't important here, because they are interested in the probability that the sample mean is within $500k of population mean. We can just assume the population mean is 0 and put the two sample means at +500k and -500k.

Given the ratio of N to n, safe to say distribution is normal.

Given standard deviation of the population at 2845000 plug everything in.

A. (-500-0)/(2845000)*(sqrt(26))

will be your left z score and

B. (500-0)/(2845000)*(sqrt(26)) will be your right z score

Take Answers from B - Answer from A and should get you your result.
 
you z scores are -0.896 and +0.896 respectively (round to 0.90 for most z tables)

gets you probabilities of 0.1841 and 0.8159

gets you 0.6318
 
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