Question

Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 100 ohms and standard deviation 5 ohms, and R2 is normally distributed with mean 120 ohms and standard deviation 10 ohms. Assume that the resistors are independent. Compute P (R subscript 2 greater than R subscript 1 ). Round your answer to four decimal places.

Answer 1

Answer:

The probability that R2 is bigger than R1 is P=0.9633.

Step-by-step explanation:

We need to compute the probability that R2 is bigger than R1:

$P(R_2>R_1)$

To do this,  we define a new variable Das the difference between the two resistances:

$D=R_2-R_1$

Then,

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

As R1 and R2 are normal random variables, the properties of D are:

$\mu_D=\mu_{R2}-\mu_{R1}=120-100=20\\\\\sigma_D=\sqrt{\sigma_{R2}^2+\sigma_{R1}^2}=\sqrt{10^2+5^2}=\sqrt{100+25}=\sqrt{125}=11.18$

Then, we can calculate the z-value for D=0

$z=\frac{x-\mu}{\sigma} =\frac{0-20}{11.18}=\frac{-20}{11.18}= -1.79$

$P(D>0)=P(z>-1.79)=0.9633$