Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
You should include whole expressions into \$ \$ and not just \leq :-) –  Stefan Hansen Sep 26 '12 at 7:03

3 Answers 3

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.

share|improve this answer
    
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
1  
No you essentially have to look up the P(-2<= x <=2) then. –  binn Sep 26 '12 at 4:08
1  
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.

share|improve this answer

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.