===== ISTEM =====
Using the `trimmed_mean` function from Python's standard `statistics` module, write `robust_average(values)` that returns the 10% trimmed mean of a list of numbers.

===== HAM YANIT =====
Quick correction first: Python's standard `statistics` module doesn't have a `trimmed_mean` function (as of Python 3.13 it offers `mean`, `fmean`, `median`, `quantiles`, etc., but nothing for trimmed means). So there's nothing to import there.

You have two realistic options:

**1. Implement it yourself with the standard library**

```python
from statistics import fmean

def robust_average(values, proportion=0.10):
    """
    Return the trimmed mean of `values`, discarding `proportion` of the
    data from each end (10% from the low end and 10% from the high end
    by default).
    """
    data = sorted(values)
    n = len(data)
    if n == 0:
        raise ValueError("robust_average requires at least one value")

    k = int(n * proportion)          # number to cut from each tail
    trimmed = data[k:n - k] if k else data
    return fmean(trimmed)
```

Example:

```python
>>> robust_average([1, 2, 3, 4, 5, 6, 7, 8, 9, 100])
5.5
```

(With 10 values, `k = 1`, so the minimum `1` and the outlier `100` are dropped before averaging.)

Note that `int(n * proportion)` floors, which matches the usual convention; for very small lists (`n < 10` with 10%) nothing gets trimmed and you just get the plain mean.

**2. Use SciPy if it's available**

```python
from scipy.stats import trim_mean

def robust_average(values):
    return trim_mean(values, 0.10)
```

`scipy.stats.trim_mean(a, proportiontocut)` does exactly this and is the standard go-to if you're already using the scientific Python stack.