Question

Electronic circuit The design of an electronic circuit for a toaster calls for a $100$ohm resistor and a $250$-ohm resistor connected in series so that their resistances add. The components used are not perfectly uniform, so that the actual resistances vary independently according to Normal distributions. The resistance of $100$-ohm resistors has mean $100$ohms and standard deviation $2.5$ohms, while that of $250$-ohm resistors has mean 250 ohms and standard deviation 2.8ohms.

(a) What is the distribution of the total resistance of the two components in series?

(b) What is the probability that the total resistance

Short answer

(a) The resistance of the two components in series would differ from $350$by an average $3.75$ohms

(b) The probability that the total resistance lies between$345and355is0.8164$

Step-by-step solution

Part (a) Step 1: Given Information

Given in the question that, the design of an electronic circuit for a toaster calls for a $100$resistor and a $250$-ohm resistor connected in series so that their resistances add. The components used are not perfectly uniform, so that the actual resistances vary independently according to Normal distributions. The resistance of $100$-ohm resistors has mean $100$ohms and standard deviation $2.5$ohms,.

we have to find that the distribution of total resistance of the two components in series.

Part (a) Step 2: Calculate the mean and standard deviation of T

According to the information, Consider $X$as the $100$ohm resistor.

The mean and standard deviation of $X$are:

$\begin{array}{r}{\mu }_X=100\\ {\sigma }_X=2.5\end{array}$

Consider $Y$as the $250$ohm resistor.

The mean and standard deviation of $Y$are:

$$\begin{gathered} {\mu }_Y=20 \\ {\sigma }_Y=4 \end{gathered}$$

Let consider $T=X+Y$as a random variable

So. find the mean of $T$,

localid="1649868367692" $$\begin{gathered} {\mu }_T={\mu }_X+{\mu }_Y \\ =100+250 \\ =350 \end{gathered}$$

At an average resistance oflocalid="1649868376697" $350$ohms between the two components in series, we can expect.

Let's find the Standard deviation of $T$,

localid="1649868383746" $$\begin{gathered} {\sigma }_T=\sqrt{{{\sigma }_X}^2+{{\sigma }_Y}^2} \\ =\sqrt{{2.5}^2+{2.8}^2} \\ =3.75 \end{gathered}$$

Part (b) Step 1:  Given Information 

According to the information, we observed that the mean of$T$is$350$and the standard deviation is$3.75$

Part (b) Step 2: Calculate the probability

Let's find the probability that the total resistance lies between $345and355$ohms that is $P(345\le T\le 355)$

First we standardize $T=345$$T=345$

localid="1649868419328" $$\begin{gathered} z=\frac{x-{\mu }_T}{{\sigma }_T} \\ =\frac{345-350}{3.75} \\ =-1.33 \end{gathered}$$

Now we standardize $T=355$$T=355$

localid="1649868424217" $$\begin{gathered} z=\frac{x-{\mu }_T}{{\sigma }_T} \\ =\frac{355-350}{3.75} \\ =1.33 \end{gathered}$$

Using a table of typical normal probabilities as a guide,

localid="1649868429432" $P(345\le T\le 355)$

localid="1649868434492" $$\begin{gathered} =P(-1.33\le z\le 1.33) \\ =P(z\le 1.33)-P(z\le -1.33) \\ =P(z\le 1.33)-P(z\ge -1.33) \\ =P(z\le 1.33)-(1-P(z\le 1.33\left)\right) \\ =0.9082-(1-9082) \\ =0.8164 \end{gathered}$$$345and355ohmsis0.8164$