Generate a random number sequence to get and average

Question

I am looking to generate a number sequence where each number is between 70 and 100 there will be x numbers in the sequence and it will give and average of y. What would this algorithm look like?

does the average of the numbers have to be exactly y or just when run many times come out to y on average? — hackartist, Jun 22 '12 at 01:28
You won't get a good answer until you tell us the distribution. That is the hard part. — John Watts, Jun 22 '12 at 01:36
Duplicate. https://stackoverflow.com/questions/39435481/how-to-generate-numbers-in-range-with-specific-average-with-python — Perennial, Feb 16 '19 at 15:48

Thilo · Accepted Answer · 2012-06-22T01:51:13.143

2

I think it is impossible for them to be uniformly distributed between 70 and 100 and have a given average at the same time.

What you can do is generate random numbers that have a given average and then scale them to fit into [70, 100] (but they will not be uniformly distributed there).

generate random numbers [0..1(
calculate their average
multiply all of them to match the required average
if any of them does not fit into [70, 100], scale all of them again by reducing their distance from y by the same factor (this does not change the average). x[i] = y + (x[i] - y)*scale

You will end up with numbers that are all in the range [70, 100(, but they will be uniformly distributed across a different (but overlapping) interval that is centered on y. Also, this approach only works with real/floating-point numbers. If you want integers, you got a combinational problem on your hands.

edited Jun 22 '12 at 01:51

answered Jun 22 '12 at 01:35

Thilo

257,207
101
511
656

right, they can't be uniformly distributed. If y was 72 for example then the numbers between 70 and 72 must appear much more often than the numbers in 72 to 100 for them to still average to 72. – hackartist Jun 22 '12 at 01:42
which is still *possible* to happen in a uniform distribution, but *extremely* unlikely. Uniform distribution implies that there are *no* restrictions on how the numbers are chosen. – Thilo Jun 22 '12 at 01:43
by the way: do we want integers? Because if we do, this becomes a combinatory problem. My solution applies only for real numbers. – Thilo Jun 22 '12 at 01:45

score 0 · Answer 2 · answered Jun 22 '12 at 01:30

0

Python example

import random
import time

x     = 10
total = 0
avg   = 0


random.seed(time.time())
for x in range(10):
    total += random.randint(70,100)

avg = total /x

print "total: ", total
print "avg: ", avg

answered Jun 22 '12 at 01:30

8bitwide

2,071
1
17
24

but here you want the "avg" to come out to a fixed number specified upfront. – Thilo Jun 22 '12 at 01:31

score 0 · Answer 3 · answered Jun 22 '12 at 01:39

        Random r = new Random();
        List<int> l = new List<int>();
        Console.Write("Please enter amount of randoms ");
        int num = (int)Console.Read();
        for (int i = 0; i < num; i++)
        {
            l.Add(r.Next(0, 30) + 70);
        }

        //calculate avg
        int sum = 0;
        foreach (int i in l)
        {
            sum += i;
        }

        Console.Write("The average of " + num + " random numbers is " + (sum / num));

        //to stop the program from closing automatically
        Console.ReadKey();

no, part of the question was that the average must be specified before hand not found afterwards. — hackartist, Jun 22 '12 at 01:40

Generate a random number sequence to get and average

3 Answers3

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