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I'm new here and was hoping you guys could help me with a statistics problem that I don't quite understand. I'm not sure if it's proper etiquette to ask for help on a specific homework problem here, so I apologize if this question is out of line.

Here is the problem:

Suppose a variable of a population has mean 5 and standard deviation 11. For samples of size 121, find $c$ so that $P(X > 2c) = 0.3300$.

Thank you

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  • $\begingroup$ Hint 1. There is probably an implicit or explicit statement that the variable is drawn from a normal or Gaussian distribution or that you can use the Central Limit Theorem to assume that the mean of a variable with finite variance is approximately normally distributed. $\endgroup$ – Henry Apr 5 '11 at 1:28
  • $\begingroup$ Hint 2. You have probably been taught how to find the variance of the mean (or perhaps sum) of $n$ independent identically distributed random variables. $\endgroup$ – Henry Apr 5 '11 at 1:30
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P(x>2c) = P(z>(2c-5)/(11/11)) = P(z>(2c-5)) = 1-P(z<(2c-5)).

Using the inverse normal function, we find that Z(.44) = .6700 which implies that Z(-.44) = .33.

THus we can set 2c-5 = -.44 then 2c = 4.56 and thus, c = 2.28.

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