About what percent of$x$ values lie between the mean and one standard deviation?

Given in the question that, We have to find the percent of $x$ values lie between the mean and one standard deviation.

We will use the following:

Let variable $\mathrm{X}$has normal distribution with mean $\mu$and variance ${\sigma }^2$. Then probability density function is defined by:

$f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{1}{2}\frac{(x-\mu {)}^2}{{\sigma }^2}},x\in R$

From famous rule in normal distribution we have following:

$P(\mu -\sigma \le X\le \mu +\sigma )=0.6827,$

$P(\mu -2\sigma \le X\le \mu +2\sigma )=0.9545,$

$P(\mu -3\sigma \le X\le \mu +3\sigma )=0.9973$.

First probability declares that $68.27\%$of $x$values lie between one-sigma interval to the left and the right of the mean. Second probability declares that $95.45\%$of $x$values lie between two-sigma intervals to the left and the right of the mean. The third probability declares that $99.73\%$of $x$values lie between the three-sigma interval to the left and the right of the mean.

We understand from earlier information that in the interval between one standard deviation to the left and one standard deviation to the right lie $68.27\%$ of $x$ values. Since this distribution is symmetrical it means that $34.14\%$ of the $x$ values lies on each side of the mean, e.c. $34.14\%$ of the data are one standard deviation to the left of the mean and $34.14\%$ of the data are one standard deviation to the right of the mean. So, between mean and one standard deviation, regardless of which side of the mean, lie $34.14\%$ of the $x$ values.