Finding given probability

Question

We are given a sample mean $\bar x=15$, standard deviation $s.d=9$. The problem is to find the percentage of data lies between $6$ and $24$ assuming that the data distribution if fairly symmetric. (Options: $68\%,81.5\%,95\%,99.7\%$).

According to me the answer should be: $P(6<X<24)$, which is obtained as follows:

$P(6<X<24)=P(-1<\frac{X-\bar x}{s.d}<1)$.

Here, $\frac{X-\bar x}{s.d}$ can be taken approximately to be $N(0,1)$ since $X$ is fairly symmetric, so the normal approximation would be good enough. Thus,

$P(-1<\frac{X-\bar x}{s.d}<1)=P(-1<Z<1)$

But to solve this, at least $P(Z<1)$ would be required (according to me!). But in the question no such info is available. Am I missing something. I thought of chebyshev's inequality but that would only give lower/upper bound of the probability. Can anyone help ?

Answer 1

If this was on a test or homework, it might just be expected that you have memorized that the amount of data within one standard deviation for a normal distribution is 68%. This could be computed using normal distribution CDF tables or similar (which would yield the $P(Z < 1)$ value you mentioned).

For an illustration see: https://en.wikipedia.org/wiki/Standard_deviation#/media/File:Standard_deviation_diagram.svg