Chapter 6.2, Problem 60E

a.

To determine

To find: The distribution of total resistance.

Explanation of Solution

Given:

Population mean $\left(\mu \right)=100$

Population standard deviation $\left(\sigma \right)=2.5$

Population mean $\left(\mu \right)=250$

Population standard deviation $\left(\sigma \right)=2.8$

Calculation:

The mean and standard deviation can be calculated as:

  $\begin{array}{c}{\mu }_{X_1+X_2}={\mu }_{X_1}+{\mu }_{X_2}\\ =100+250\\ =350\end{array}$

  $\begin{array}{c}{\sigma }_{X_1+X_2}=\sqrt{{\sigma }^2{}_{X_1}+{\sigma }^2{}_{X_2}}\\ =\sqrt{{2.5}^2+{2.8}^2}\\ =3.75\end{array}$

b.

To determine

To find: The probability that total resistance falls between 345 and 355 ohms.

Answer to Problem 60E

The probability is 0.8164.

Explanation of Solution

Calculation:

The probability that total resistance falls between 345 and 355 ohms can be computed as:

  $\begin{array}{c}P\left(345+X_1+X_2<355\right)=P\left(\frac{345-350}{3.75}<Z<\frac{355-350}{3.75}\right)\\ =P\left(-1.33<Z<1.33\right)\\ =0.8164\end{array}$

Thus, the probability is 0.8164.