Solution Found!
Two resistors, with resistances R1 and R2, are connected
Chapter 4, Problem 21E(choose chapter or problem)
Two resistors, with resistances \(R_{1}\) and \(R_{2}\), are connected in series. \(R_{1}\) is normally distributed with mean 100\(\Omega\) and standard deviation 5\(\Omega\), and \(R_{2}\) is normally distributed with mean 120\(\Omega\) and standard deviation 10\(\Omega\).
a. What is the probability that \(R_{2}\) > \(R_{1}\)?
b. What is the probability that \(R_{2}\) exceeds \(R_{1}\) by more than 30\(\Omega\)?
Equation transcription:
Text transcription:
\Omega
R_{2}
R_{1}
Questions & Answers
QUESTION:
Two resistors, with resistances \(R_{1}\) and \(R_{2}\), are connected in series. \(R_{1}\) is normally distributed with mean 100\(\Omega\) and standard deviation 5\(\Omega\), and \(R_{2}\) is normally distributed with mean 120\(\Omega\) and standard deviation 10\(\Omega\).
a. What is the probability that \(R_{2}\) > \(R_{1}\)?
b. What is the probability that \(R_{2}\) exceeds \(R_{1}\) by more than 30\(\Omega\)?
Equation transcription:
Text transcription:
\Omega
R_{2}
R_{1}
ANSWER:
Answer:
Step 1 of 2:
(a)
In this question, we are asked to find the probability that
is normally distributed with mean 100 and standard deviation .
is normally distributed with mean 120 and standard deviation .
Both resistor are connected in series.
We need to find the we can rewrite as,
=
=
Where .
Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.
Since is a linear combination of independent normal random variable, is also normally distributed.
Now we will calculate the z-score, to find
The z-score of 0 is z =
From the z table, the area to the left of is 0.0375.
The area to the right of is
Therefore .