Probability

[#1]

Hello,

I have the following distribution with a mean of 0,027.

Based on this distribution, is there a way to find out what is the probability of a number to be more than or equal to 0,000269? Is there also a way to simulate how many times in a row can the number be less than 0,000269?

Thank you!

[#2]

You can use the Normal Distribution() function to get the probability answer you asked and you can use the Random Normal() function to generate a random sample which you can then count up the number of values in a row result you want.

[#3]

Thank you for your answer, but I am not sure where to find the normal distribution function.

[#4]

If you open the Formula editor in the Column header you find different options:

[#5]

I found this probability calculator and entered the mean and standard deviation of my distribution. Then I entered the value I want to observe and it gives a 50,89% probability of a number greater than that of occurring. Is that correct?

I have also tried using the method you have suggested.

However, something does not seem right to me (I think I am not doing it right).
Does that mean that there is a 50,01% probability of a number less than 0,00269 to occur?

[#6]

names default to here(1);
prob = normal distribution(0.000269, 0.0270052, 1.2004589);
double_row_prob = Prob * Prob;

This gives you the probabilty of a value beeing less than 0.000269 if the distribution is normal with the mentioned parameters.

The probability needs to be less than 50% because the value is less than the mean. In your screenshot you looked at the wrong side of the graph.

For two in a row for independent events you need to multiply the single probability.

[#7]

Thank you, but it does not make sense to me to be less than 50%. Yes, the number is less than the mean, but I am looking for the probability of the numbers equal and higher than it to occur. So, there is a 50,89 probability that a number bigger than 0.000269 will occur (using the distribution calculator I found). Because there are more numbers between this number and the last number. With your formula, it gives 49,11% but isn't that calculating only the % of the number to occur, not the range from [0.000269; last positive number]

[#8 · Accepted solution]

sorry, I misread your first sentence in your first post, your're right.

normal distribution(q, mu, sigma) gives the probability of of a value being less or equal than q when drawn from a normal distributed population with parameters mu and sigma (see manual or e.g. scripting index).

So the probability of a value being higher than q is:

1-normal distribution(q, mu, sigma).

Have a look at the scripting index, there you will find it as written below.

The normal distribution function is the integral of the normal density function you saw in the distribution calculator.

Names Default To Here( 1 );
New Window( "Example: Normal Distribution",
y = Graph Box(
Y Scale( 0, 1 ),
X Scale( -4, 4 ),
XName( "q" ),
Pen Color( "red" );
Y Function( Normal Distribution( q ), q );
)
);