math stats with Linq

var max = persons.Max(p => p.age);
var min = persons.Min(p => p.age);
var average = persons.Average(p => p.age);

Fix for median in case of even number of elements

int count = persons.Count();
var orderedPersons = persons.OrderBy(p => p.age);
float median = orderedPersons.ElementAt(count/2).age + orderedPersons.ElementAt((count-1)/2).age;
median /= 2;

Max, Min, Average are part of Linq:

int[] ints = new int[]{3,4,5};
Console.WriteLine(ints.Max());
Console.WriteLine(ints.Min());
Console.WriteLine(ints.Average());

Median is easy:

UPDATE

I have added order:

ints.OrderBy(x=>x).Skip(ints.Count()/2).First();

BEWARE

All these operations are done in a loop. For example, ints.Count() is a loop so if you already get ints.Length and stored to a variable or simply use it as it is, would be better.

Here is a complete, generic implementation of Median that properly handles empty collections and nullable types. It is LINQ-friendly in the style of Enumerable.Average, for example:

    double? medianAge = people.Median(p => p.Age);

This implementation returns null when there are no non-null values in the collection, but if you don't like the nullable return type, you could easily change it to throw an exception instead.

public static double? Median<TColl, TValue>(
    this IEnumerable<TColl> source,
    Func<TColl, TValue>     selector)
{
    return source.Select<TColl, TValue>(selector).Median();
}

public static double? Median<T>(
    this IEnumerable<T> source)
{
    if(Nullable.GetUnderlyingType(typeof(T)) != null)
        source = source.Where(x => x != null);

    int count = source.Count();
    if(count == 0)
        return null;

    source = source.OrderBy(n => n);

    int midpoint = count / 2;
    if(count % 2 == 0)
        return (Convert.ToDouble(source.ElementAt(midpoint - 1)) + Convert.ToDouble(source.ElementAt(midpoint))) / 2.0;
    else
        return Convert.ToDouble(source.ElementAt(midpoint));
}