Equation with normal distribution function

Question

I was working on a task in probability, and got stuck at this:

$ϕ(\frac{x-50}{4}) - ϕ(\frac{-x-50}{4}) = 0.6$

($ϕ$ is the normal distribution function.)

It's so simple, yet I don't know what to do about it.

Edit:

The task is: find a symmetric interval which contains $60$% of cases, if the random variable has normal distribution with parameters ($50$, $16$).

Edit No.2

@dmk:

I actually don't know what a z-score is (haven't learned that or at least it's not called that here), but I solved the task like this and got the correct result:

Since the distribution is symmetric around $50$, my interval has to range from $50-x$ to $50+x$. So, the interval from $50-x$ to $50$ would contain $30$% of values, and that's how I can calculate $x$:

$ϕ(0) - ϕ(\frac{50-x-50}{4}) = 0.3$

which gives $x=3.4$.

Answer 1

Since the normal distribution is symmetric, what the question is essentially asking is: Can you find a $z$-score such that the area between $0$ and $z$ is $0.6 / 2 = 0.3$? Since a $z$-score is for a standardized normal distribution, you'll have to then convert this back to unstandardized form by setting $z$ equal to one of the given inputs for the cdf; note that the input is of the form $\frac{x-\mu}{\theta}$.

What value of $x$ do you get? (Let me know if you have any questions.)

Edit: So, from your second edit, it looks as though you're trying to work with the left half of the interval $[50-x, 50+x]$ — namely, you need the area between $50-x$ and $50$ to be $0.3$. I'm going to use the right half, just because it works out a little more cleanly. So the equality toward the end of your edit should look like this:

$$ \begin{align} \ \Phi\left(\frac{50+x}{4}\right) - \Phi\left(\frac{50-50}{4}\right) &= 0.3 \\ \ \Phi\left(\frac{50+x}{4}\right) - \Phi(0) &= 0.3 \\ \ \Phi\left(\frac{50+x}{4}\right) - 0.5 &= 0.3 \\ \ \Phi\left(\frac{50+x}{4}\right) &= 0.8 \\ \end{align} $$

At this point, you need some method of starting with an answer (i.e., the $0.8$) and working out the question (i.e., the input of the cdf, $\Phi$). Usually, this is done with a $z$ table. The $z$ table tells us that $\Phi(0.84) \approx 0.8$. So then we let $0.84 = \frac{50+x}{4}$, from which it follows that $x = 3.36$. Thus, the endpoints of the interval are $50 \pm 3.36$, which gives us the required interval length.

Answer 2

For a standard normal distribution, you should use $\Phi$ for the cumulative distribution function, and $\phi$ for the density. I also suspect you want $-\frac{x-\mu}{\sigma}$ trather than $\frac{-x-\mu}{\sigma}$. Making both changes will give you the answer in the book, or close enough.

Since a normal distribution $N(\mu,\sigma^2)$ is symmetric about the mean, you will have $$\Phi\left(-\dfrac{x-\mu}{\sigma}\right) = 1- \Phi\left(\dfrac{x-\mu}{\sigma}\right)$$ or alternatively (essentially the same result with a different approach) $$ \Phi\left(\dfrac{x-\mu}{\sigma}\right) -\Phi\left(-\dfrac{x-\mu}{\sigma}\right) = 2\left(\Phi\left(\dfrac{x-\mu}{\sigma}\right) -\Phi\left(0\right)\right).$$ With the information in your question, that will allow you to find $\Phi\left(\dfrac{x-\mu}{\sigma}\right)$ and thus $\dfrac{x-\mu}{\sigma}$ and $x$, and so the difference between the upper and lower points of the interval which will be $x- (\mu - (x-\mu))$.