Chapter 6.2, Problem 66E

(a)

To determine

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

Answer to Problem 66E

The distribution of the total resistance of the two components in a series is Normal distribution.

Explanation of Solution

Given information:

X : Resistance of a 100 − ohm resistor

Y : Resistance of a 250 − ohm resistor

For X :

Mean,

  ${\mu }_X=100\text{ohms}$

Standard deviation,

  ${\sigma }_X=2.5\text{ohms}$

For Y :

Mean,

  ${\mu }_Y=250\text{ohms}$

Standard deviation,

  ${\sigma }_Y=2.8\text{ohms}$

X and Y are the two components in series.

Thus,

X + Yis the total resistance of two components in series.

Both X and Y has a normal distribution.

That means

X + Y also have a normal distribution.

If X and Y are independent,

Property mean:

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

Property variance:

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

Thus,

We have

Mean of total resistanceX + Y ,

  ${\mu }_{X+Y}={\mu }_X+{\mu }_Y=100+250=350\text{ohms}$

Variance of total resistance X + Y ,

  ${\sigma }_{X+Y}^2={\sigma }_X^2+{\sigma }_Y^2={\left(2.5\right)}^2+{\left(2.8\right)}^2=14.09\text{ohms}$

We know that

Standard deviation is the square root of the variance.

Standard deviation of total resistanceX + Y ,

  ${\sigma }_{X+Y}=\sqrt{{\sigma }_X^2+{\sigma }_Y^2}=\sqrt{14.09}=\sqrt{20}\approx 3.7537\text{ohms}$

Thus,

The distribution of total resistance X + Y will be the Normal distribution.

(b)

To determine

Probability for the total resistance for a randomly selected toaster lies between 345 and 355 ohms.

Answer to Problem 66E

Probability that the total resistance for a randomly selected toaster lies between 345 and 355 ohms is 0.8164.

Explanation of Solution

Given information:

  $x=X+Y=345\text{ohms}\text{or}\text{355}\text{ohms}$

From Part (a),

Mean,

  $\mu =350\text{ohms}$

Standard deviation,

  $\sigma =3.7537\text{ohms}$

Calculations:

Calculate the z − score,

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

Or

  $z=\frac{x-\mu }{\sigma }=\frac{355-350}{3.7537}\approx 1.33$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.3 and the column that starts with .03 of the standard normal probability table for $P\left(z<-1.33\right)$ .

Or

See the row that starts with 1.3 and the column that starts with .03 of the standard normal probability table for $P\left(z<1.33\right)$ .

  $\begin{array}{l}P\left(345<X+Y<355\right)=P\left(-1.33<z<1.33\right)\\ =P\left(z<1.33\right)-P\left(z<-1.33\right)\\ =0.9082-0.0918\\ =0.8164\end{array}$

Thus,

The probability that the total resistanceX + Y lies between 345 and 355 ohms is 0.8164.