Question

Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 100Ω and standard deviation 6Ω, and R2 is normally distributed with mean 120Ω and standard deviation 10Ω. What is the probability that R2 > R1?

Answer 1

Answer:

$P(R_2 -R_1 >0) = P(d>0) = P(z>\frac{0-20}{\sqrt{136}}) = P(z>-1.715)$

And we can find this probability using the complement rule and with the normal standard distribution or excel like this:

$P(z>-1.715) = 1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

For this case we have two resistors with the following distributions:

$R_1 \sim N (\mu_1 =100, \sigma_1 = 6)$

$R_2 \sim N (\mu_2 =120, \sigma_2 = 10)$

And for this case we want the following probability:

$P(R_2 >R_1) = P(R_2 -R_1 >0)$

First we need to find the distributon for $R_2- R_1$

Since both are normally distributed we have that a linear combination of normal distributions is also normal, so then we have this:

$d=R_2 -R_1 \sim N (\mu_d=\mu_2-\mu_1 = 120-100= 20, \sigma^2_d=\sigma^2_1 +\sigma^2_2= 36+100=136)$

And we can use the z score formula given by:

$z= \frac{d-\mu_d}{\sigma_d}$

And using this formula we got:

$P(R_2 -R_1 >0) = P(d>0) = P(z>\frac{0-20}{\sqrt{136}}) = P(z>-1.715)$

And we can find this probability using the complement rule and with the normal standard distribution or excel like this:

$P(z>-1.715) = 1-P(Z$