Solving Probability Problems: Normal Distributions and the Chance of B < A

[Marije]

Probability question

I can't figure out how to solve this problem.

Given are two normal distributions, A: Norm(12; 0,4) and B: Norm(13; 0,9).
If we pick one from A and one from B what's the chance B < A?
I don't know how to do this because we haven't learned a formula or anything for the graphs. I think what they're asking for is to calculate the area I coloured in this picture:
http://img211.imageshack.us/img211/1692/kansberekeningsk5.jpg
Sorry for the crappy drawing skills. I hope you understand the idea anyway.
Is there any way to calculate it? ;x I'm lost.

Oops sorry I think I posted this in the wrong section... Sorry, I'm new here. I'm not sure if this should go here or in the homework section, since there are more questions like mine in this forum but it's homework so I don't know, sorry.

 

[mathman]

Have you learned how to get the distribution for the sum (or difference) of two independent random variables, particularly of normal variables? If so, let x be chosen from A, y chosen from B, and let z=x-y. z will be normally distributed and you need to get prob (z<0).

 

[tacman]

Hi there,

No worries, this is the right section to post your question. Solving probability problems involving normal distributions can be tricky, but there are certain steps you can follow to find the solution.

First, let's define the problem in mathematical terms. We have two normal distributions, A and B, with means of 12 and 13, and standard deviations of 0.4 and 0.9, respectively. We want to find the probability that a randomly selected value from distribution B is less than a randomly selected value from distribution A.

To solve this problem, we need to use the concept of z-scores. A z-score is a measure of how many standard deviations a value is away from the mean. It is calculated by subtracting the mean from the value and then dividing by the standard deviation.

In this case, we want to find the probability that B < A, so we need to calculate the z-score for the value of 13 (mean of distribution B) in relation to distribution A. This can be done by using the formula z = (x - μ)/σ, where x is the value we want to find the z-score for, μ is the mean, and σ is the standard deviation. Plugging in the values, we get z = (13 - 12)/0.4 = 2.5.

Next, we need to find the corresponding probability for this z-score. This can be done by using a z-score table or a calculator. For a z-score of 2.5, the probability is approximately 0.9938.

Now, we need to find the probability that B > A. This is simply 1 - 0.9938 = 0.0062.

Therefore, the probability that B < A is 0.0062. In other words, there is a very low chance that a randomly selected value from distribution B will be less than a randomly selected value from distribution A.

I hope this helps you understand how to approach and solve this type of probability problem. If you have any further questions, feel free to ask.

Best of luck with your studies!