Suppose we have data as shown in the histogram below. (Of course there are no ages below $0,$ but I wanted the histogram bars to be of equal widths.)
Suppose we make the simplifying approximation that all ages in each of the $k = 4$ histogram bins are at the centers of their intervals. Then we get frequencies $3, 41, 51, 5$ (for $n = 100$ subjects altogether) 'located' at midpoints $7, 25, 45, 65.$
Then the sample mean can be estimated as $$\bar X \approx \frac{1}{n}\sum_{i = 1}^k f_im_i = 36.66.$$
The average was computed, using R statistical software as a calculator, as follows:
f = c(3, 41, 51, 5)
m = c(7, 25, 45, 65)
a = sum(f*m)/100; a
[1] 36.66
Perhaps somewhat less accurately, we can approximate the sample variance as follows:
$$S^2 \approx \frac{1}{n-1}\sum_{i=1}^k f_i(m_i - \bar X)^2 = 159.358.$$
v = sum(f*(m-a)^2)/99; v
[1] 159.358
Unless you have the original data, you can't know how accurately $\bar X$ and $S^2$ are actually estimated by these formulas. However, I simulated the heights in R, so we can check the true values. For my simulated data the approximate values happen to be quite accurate. [I suspect results in this example turned out to be a little better than usual--especially because we used only four intervals.]
set.seed(2020) # for reproducibility
x = round(rnorm(100, 35, 10))
summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
5.00 29.00 36.00 36.09 42.25 67.00
var(x)
[1] 124.4666
Note: (1) Some years ago when computation was more tedious than it is with modern software, it was common practice to use the formulas shown here to approximate $\bar X$ and $S^2$ for large samples. Results are usually pretty good if you use a dozen or so intervals.
(2) Some elementary statistics books (especially high-school AP statistics books) have formulas for estimating sample medians, other sample quantiles, estimating population modes from data summarized in the form of intervals and frequencies.
(3) See this related Q&A.