Normal Random Variable

Question

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?

i'm pretty sure the answer is pnorm(2) - pnorm(-2) where pnorm is the standard normal cdf — binn, Sep 26 '12 at 03:17
You should include whole expressions into $ $ and not just \leq :-) — Stefan Hansen, Sep 26 '12 at 07:03

score 4 · Answer 1 · answered Sep 26 '12 at 04:15

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.

score 2 · Accepted Answer · answered Sep 26 '12 at 03:23

2

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.

answered Sep 26 '12 at 03:23

Bitwise

1,216

So the answer will be 97.7 ? – Q.matin Sep 26 '12 at 03:26
97.7 is not a probability... – Bitwise Sep 26 '12 at 03:27
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 03:34
hmm will it be (1 - .977) then for both sides ? – Q.matin Sep 26 '12 at 03:38
Think about it this way: 0.977 is the probability of X<=106 . So the probability of the X>106 is (1-0.977). This covers only half of the cases you want. In the other half, X<94 with the same probability. So what will the final probability for 94<=X<=106 be? – Bitwise Sep 26 '12 at 03:45
Wait so I essentially have to look up the P(1<= x <=2) then? So it will be .9772 - .8413 = 0.1359 ? – Q.matin Sep 26 '12 at 03:46
1

No you essentially have to look up the P(-2<= x <=2) then. – binn Sep 26 '12 at 04:08
1

1-2*(1-0.977) ... – Bitwise Sep 26 '12 at 04:37

score 0 · Answer 3 · answered Oct 17 '13 at 20:26

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).

Normal Random Variable

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