How to generate a series of numbers based on an average

average

I'm not too good with math so I'm struggling with this. Say I am given an average yearly temperature and I want to generate twelve random numbers to constitute the average for each month of the year within a certain limit so that I'm not getting temperatures that are unrealistically high. In other words, generating a series of twelve numbers from an already given average, and those twelve numbers equal the average. Does that make sense? I hope it does. Like I said, I'm not too good with math. Thanks for any feedback!

Best Answer

Choose a tight interval of reasonable average temperatures for each month at your location. For example, in Boston for January that might be $(10 ,30)$ in degrees Fahrenheit.

Choose a random value $V$ in each of those $12$ intervals.

Find their average $A$.

Calculate the difference $d = A-B$ between $A$ and the average $B$ you want.

Subtract $d$ from each month's $V$.

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[Math] Algorithm to separate an array of numbers into “low”, “medium” and “high” ranges

It's a little bit unclear what you want, so I'm going to present several possible answers that you can choose from.

The simplest possibility is that you simply want to divide the range into three equal parts. Assuming that the lowest number is $L$ and the highest is $H$, you can calculate the ranges as:
1. from $L$ to $L + \frac{H-L}3$,
2. from $L + \frac{H-L}3$ to $L + 2\frac{H-L}3$ and
3. from $L + 2\frac{H-L}3$ to $H$.
In fact, a simple way to assign a number $n$ into one of these ranges is to subtract $L$ from it, multiply the result by $3$, divide it by $H-L$ and truncate it to the next lowest integer:
```
group = int( (n - L) * 3 / (H - L) );
```
This will give you a number from $0$ to $2$, or possibly $3$ if $n = H$. To avoid that last case, you can either replace $H$ in the formula with $H+1$ (if working with integers) or $H + \epsilon$ for some very small number $\epsilon$ (if working with floats), or just check for it afterwards, e.g.:
```
if (group > 2) group = 2;
```
Obviously, you can use the exact same method for more groups than 3 as well.
Another possibility is that you want to divide the numbers into three groups of equal size, such that all numbers in group 1 are less than or equal to those in group 2, which are less than or equal to those in group 3. In other words, you want to divide your numbers into three quantiles.

The simplest way to do this programmatically is just to sort your list of numbers, and then split the sorted list into three parts of (approximately) equal length. In pseudocode, using Python-like array slice notation:
```
list.sort();
n = list.length;
part1 = list[0 : n/3];
part2 = list[n/3 : 2*n/3];
part3 = list[2*n/3 : n];
```

[Math] Average time between successive occurrences of earthquakes

Let $X$ be the number of years to the next earthquake.

Then $X$ is a geometric random variable and the

probability that the next earthquake happens in $k$ years is

$P(X = k) = (1-p)^k \cdot p$

where $p = 0.04$.

The expected value of $X$ is

$E(X) = 1/p = 1/0.04 = 25 \:\text{years.}$

And this can also be interpreted as if you waited a long period of time the average time between earthquakes would be $25$ years.

(or you can think of it that the distribution of the length of the gaps between earthquakes will follow a geometric distribution)

Best Answer

Related Solutions

[Math] Algorithm to separate an array of numbers into “low”, “medium” and “high” ranges

[Math] Average time between successive occurrences of earthquakes

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