Two resistors, with resistances R1 and R2, are connected

Chapter 4, Problem 21E

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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}

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 .


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