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I'm working on an excercise in which I have problem understanding if I should assume it's discrete or continuous random variable.

The problem states: "Given a population with mean, variance and sample size n equal to $$\mu=100,\ \sigma^2=900,\ n=30$$, what is the probability that $$P(96 <= \bar x <=110)$$

I know how to solve the calculating part and use the Central Limit Theorem, however before that I struggle to understand how I should divide it up depending on the type:

Discrete: $$P(96 <= \bar x <=110) = P(\bar x <= 110) - P(\bar x <= 95)$$

Continuous: $$P(96 <= \bar x <=110) = P(\bar x <= 110) - P(\bar x <= 96)$$

How would I know if it's discrete and therefore use (95) instead of (96)?

Looking at the solution they've used (96) and I assume they think it's continuous.

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If the samples take on discrete values, then a continuity correction is usually applied to get a better approximation. So, you should use something like $$P\{96 \leq \bar{x} \leq 110\} \approx P\{95.5 \leq Z \leq 110.5\}$$ where $Z$ is a normal random variable with the appropriate mean and variance. –  Dilip Sarwate Feb 8 '12 at 11:58
    
Yes, agree but as you wrote, it would be in the next step. –  samuelf Feb 8 '12 at 12:15
    
No, it is not in the next step. Unless you are told that $\bar{x}$ is always an integer (a much stronger condition than saying that the samples are discrete-valued), your assertion that $$P\{96 \leq \bar{x} \leq 110\} = P\{\bar{x} \leq 110\} - P\{\bar{x} \leq 95\}$$ is incorrect; it should be $$P\{96 \leq \bar{x} \leq 110\} = P\{\bar{x} \leq 110\} - P\{\bar{x} < 96\}$$ to account for the possibility that $\bar{x}$ might take on values between $95$ and $96$ with nonzero probability. –  Dilip Sarwate Feb 8 '12 at 12:29
    
Ah, I see. So you mean, because there is no assertion in the text, I cannot just assume that possibility but have to account for the possibility that it might be? –  samuelf Feb 8 '12 at 12:55
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