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I am wondering if I did this question right:

Assume that $X$ is a normal random variable. Assume that the expectation is $E[X] = 100$ and the standard deviation is 3. Determine $$ P(E[X] - 6 \leq X \leq E[X] + 6). $$

My attempt:

Since we know $E[X]$ then $P(100 - 6 \leq X \leq 100 + 6) = P(94 \leq X \leq 106)$ and from this we can look at the normal distribution table for the probablity, but isn't there a limit for the distribution table?

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i'm pretty sure the answer is pnorm(2) - pnorm(-2) where pnorm is the standard normal cdf – binn Sep 26 '12 at 3:17
@binn I dont quite understand those notations? – Q.matin Sep 26 '12 at 3:19
You should include whole expressions into \$ \$ and not just \leq :-) – Stefan Hansen Sep 26 '12 at 7:03
up vote 2 down vote accepted

What matters here is only the standard deviation. You are asked what is the probability of x being at most 2 standard deviations away from the mean (since 2*3=6). So you simply need to look up this probability in the normal distribution table (there the standard deviation will be 1, so look at stdev=2). It should be somewhere around 0.95 if I am not mistaken.

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So the answer will be 97.7 ? – Q.matin Sep 26 '12 at 3:26
97.7 is not a probability... – Bitwise Sep 26 '12 at 3:27
You are almost there... 0.977 is for one side of the distribution, how much will it be for both sides? – Bitwise Sep 26 '12 at 3:34
No you essentially have to look up the P(-2<= x <=2) then. – binn Sep 26 '12 at 4:08
1-2*(1-0.977) ... – Bitwise Sep 26 '12 at 4:37

Typical normal distribution tables give values of $\Phi(x) = P\{X \leq x\}$ for nonnegative values of $x$ where $X$ is a standard normal random variable, usually for $x$ in the range from $0$ to $3.5$. Now, for any normal random variable $Y$ with mean $\mu$ and standard deviation $\sigma$,

$$ P\{Y \leq y\} = \Phi\left(\frac{y-\mu}{\sigma}\right)$$

Remember the argument on the right as distance of $y$ from the mean $\mu$ measured in units of the standard deviation $\sigma$.

More generally, $$P\{y_1 \leq Y \leq y_2\} = P\{Y \leq y_2\} - P\{Y \leq y_1\} = \Phi\left(\frac{y_2-\mu}{\sigma}\right) - \Phi\left(\frac{y_1-\mu}{\sigma}\right)$$

The table for $\Phi(x)$ does not list values for $x < 0$ because these values can always be deduced via the relationship

$$\Phi(-x) = 1 - \Phi(x).$$

Thus, $\Phi(-1) = 1 - \Phi(1)$. So, for any normal random variable, express the probability you want to find in terms of $\Phi(x)$, and then look up values for $\Phi(x)$ in the table if $x \geq 0$, and use $\Phi(x) = 1 - \Phi(|x|)$ if $x < 0$.

Finally, if all else fails, use a calculator such as the one here to check your answer.

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Standardize it by subtracting the mean and dividing by the standard deviation. Then you have P{-2 ≤ z ≤ 2}, where z~N(0,1). Then just look up these values on a standard normal table and you'll find P{-2 ≤ z ≤ 2} = 0.9544 which is approximately 95% (2 standard deviations from the mean in either direction is the middle 95% of the data under the empirical rule).

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