Probability from the tolerance of a capacitor (Gaussian distribution)

[Peter Alexander]

Homework Statement

Given a capacitor with 33 nF
Task requires you to compute the probability for a capacitance being greater than 30nF, given that there's 20% tolerance (3σ).

Relevant Equations

The formula for Gaussian distribution (https://en.wikipedia.org/wiki/Normal_distribution)

Given the upper data, if the nominal value for capacitance is 33nF and tolerance of 20%, then values can range between 26.4nF and 39.6nF. With the bottom margin being set at 30nF, this means that the interval takes approximately 72% of all values.

Is this the correct procedure to solve this task?

Any sort of help would be appreciated.

 

[BvU]

No. How did you determine the 0.72 ?

 

[Peter Alexander]

No. How did you determine the 0.72 ?

By dividing the interval 30 - 39.6 from 26.4 - 39.6.

 

[WWGD]

If you find the z-value associated with 30, meaning the number of ##\sigma## from the expected value, you can just look up the associated percent/percentile in a standard normal table. Edit: I am assuming from your post that the data in question are normally-distributed. Please let me know if that is not correct or must be proven first.

 

[BvU]

By dividing the interval 30 - 39.6 from 26.4 - 39.6.

It would be clearer if you posted something like ' I did (39.6-30)/(39.6-26.4) = 0.727 '

Which is definitely not the idea of this exercise.

Another notion that needs correction is

values can range between 26.4nF and 39.6nF

because the exercise text clearly implies that the capacitance is distributed according to a normal distribution with average 33 nF and a standard deviation of 6.6/3 nF = 2.2 nF.
That means capacitances can range between ##-\infty## and ##+\infty##
(not to worry, the probabilities decrease very rapidly outside reasonable ranges. But theoretically they are not zero !)
Just a consequence of the assumed probability distribution model -- for which very good but not perfect arguments exist.

In fact, outside the range average ##\pm 3\sigma##, 0.27% of the values are theoretically expected.

Now, what are you supposed to do: given the average value of 33 nF and the standard deviation of 2.2 nF

compute the probability for a capacitance being greater than 30nF

Suppose you have a standard normal distribution plot in front of you ,

the probability to find any value corresponds to the total area under the curve: 1 (or also expressed as 100%)
the probability to find a value > 33nF corresponds to the area under the curve from 33 nF to infinity: 0.5 (from symmetry)

Can you describe what area corresponds to the probablity the exercise asks for ?------------------------------------

Another important bit of wise-guy comment:

What we casually call probablility distributions are actually plots of probability densities . Probabilities emerge when we multiply with a range: probability for a value to be in ##[x, x+dx]## is equal to ##P(x)\, dx##.​