The design of an electronic circuit for a toaster calls for a $100-\mathrm{ohm}$resistor and a $250-\mathrm{ohm}$resistor connected in series so that their resistances add. The resistance data-custom-editor="chemistry" $\mathrm{X}\mathrm{of}\mathrm{a}100-\mathrm{ohm}$resistor in a randomly selected toaster follows a Normal distribution with mean data-custom-editor="chemistry" $100\mathrm{ohms}$and standard deviation data-custom-editor="chemistry" $2.5\mathrm{ohms}$. The resistance data-custom-editor="chemistry" $\mathrm{Y}\mathrm{of}\mathrm{a}250-\mathrm{ohm}$resistor in a randomly selected toaster follows a Normal distribution with mean $250\mathrm{ohms}$and standard deviationdata-custom-editor="chemistry" $2.8\mathrm{ohms}$. The resistances data-custom-editor="chemistry" $\mathrm{X}\mathrm{and}\mathrm{Y}$are independent.

a. Describe the distribution of the total resistance of the two components in series for a randomly selected toaster.

b. Find the probability that the total resistance for a randomly selected toaster lies between$345\mathrm{and}355$ ohms.

Given :

$X$: Resistance of a 100-ohm resistor

$Y$: Resistance of a 250 -ohm resistor

Mean,

${\mu }_X=100$ohms

${\mu }_Y=250$ohms

Standard deviation,

${\sigma }_X=2.5$ohms

${\sigma }_Y=2.8$ohms

The two components in series are $\mathrm{X}\mathrm{and}\mathrm{Y}.$

As a result, the total resistance of two components in series is $X+Y$.

A normal distribution exists for both $\mathrm{X}\mathrm{and}\mathrm{Y}$.

As a result, $X+Y$has a normal distribution as well.

If $\mathrm{X}\mathrm{and}\mathrm{Y}$ are unrelated, Property mean :

${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

Property variance : ${{\sigma }^2}_{aX+bY}=a^2{\mu }_X+b^2{\mu }_Y$

As a result, Total resistance average :

$$\begin{gathered} {\mu }_{X+Y}={\mu }_X+{\mu }_Y \\ =100+250 \\ =350\mathrm{ohms} \end{gathered}$$

Total resistance variation :

$$\begin{gathered} {{\sigma }^2}_{x+y}={{\sigma }^2}_X+{{\sigma }^2}_Y \\ ={(2.5)}^2+{(2.8)}^2 \\ =\mathbf{14}\mathbf{.}\mathbf{09}\mathrm{ohms} \end{gathered}$$

We know that the square root of the variance is the standard deviation.

The standard deviation of total resistance $X+Y$:

$$\begin{gathered} {\sigma }_{x+y}=\sqrt{{{\sigma }^2}_X+{{\sigma }^2}_Y} \\ =\sqrt{14.09} \\ \approx \mathbf{3}\mathbf{.}\mathbf{7537}ohms \end{gathered}$$

Given :

$X$: Resistance of a $100-$ohm resistor

$Y$: Resistance of a $250-$ohm resistor

Mean,

${\mu }_X=100$ohms

${\mu }_Y=250$ohms

Standard deviation,

${\sigma }_X=2.5$ohms

${\sigma }_Y=2.8$ohms

Calculate the $z-$score using the formula

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{345-350}{3.7537} \\ \approx -1.33 \end{gathered}$$

Or

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{350-345}{3.7537} \\ \approx 1.33 \end{gathered}$$

To find the equivalent probability, use the normal probability table in the appendix. In the typical normal probability table for $P(Z<-1.33)$, look for the row that starts with $-1.3$and the column that starts with $.03$.

Or In the typical normal probability table for $P(Z<1.33)$, look for the row that starts with $1.3$and the column that starts with $.03$.

$$\begin{gathered} P(345<X+Y<355)=P(-1.33<z<1.33) \\ =P(Z<1.33)-P(Z<-1.33) \\ =0.9082-0.0918 \\ =\mathbf{0}\mathbf{.}\mathbf{8164} \end{gathered}$$