Suppose the following data are obtained by recording $X$, the number of customers that arrive at an automatic banking machine during $15$ successive one-minute time intervals.
Q) Record the mean and variance.
mean is
$u_{X} = \sum_{x=1}^{15} x f_{X}(x) = 1.67$
using the data from below:
$f_{X}(0) = 4/15, f_{X}(1) = 3/15, f_{X}(2) = 4/15, f_{X}(3) = 2/15, f_{X}(4) = 2/15$
variance is
$\sigma^2_{X} = \sum_{x=1}^{15} x f_{X}(x) - 1.67^2 = 1.88$
the mean is the same as the solution, but the variance they got was $1.952$. What am I doing wrong?
Your method finds the population variance where
$$\begin{align*} \mathsf{Var}(X) &=\mathsf E(X^2)-\mathsf E(X)^2\\\\ &=0^2\left(\frac{4}{15}\right)+1^2\left(\frac{3}{15}\right)+2^2\left(\frac{4}{15}\right)+3^2\left(\frac{2}{15}\right)+4^2\left(\frac{2}{15}\right)-\left(\frac{25}{15}\right)^2\\\\ &=4.6-\left(\frac{25}{15}\right)^2\\\\ &\approx1.882 \end{align*}$$
The sample variance, which is desired, is defined as
$$s^2=\frac{1}{n-1}\cdot\sum_{i=1}^n (X_i-\bar{X})^2$$
so we get
$$\begin{align*} s^2 &=\frac{1}{14}\left(\left(2-\frac{25}{15}\right)^2+\cdots+\left(4-\frac{25}{15}\right)^2\right)\\\\ &\approx 1.952 \end{align*}$$