Chapter 6, Problem 4CRE

(a)

To determine

To explain: The reason behind assuming that the cap strength and torque applied by machine is independent.

Explanation of Solution

Given:

T denotes the capping machine torque

C denotes the cap strength

Both of them follow normal distribution

Mean of T $=7$

Standard deviation of T $=0.9$

Mean of C $=10$

Standard deviation of C $=1.2$

Formula used:

When two events are independent

  $\begin{array}{l}E\left[A-B\right]=E\left[A\right]-E\left[B\right]\\ Var\left[A-B\right]=Var\left[A\right]+V\left[B\right]\end{array}$

Both the torque capping machine and caps on containers are different machines and strength applied are working independently.

So, both of them are working independently.

(b)

To determine

To find: The mean and standard deviation of D

Answer to Problem 4CRE

  $\begin{array}{l}E\left[D\right]=3\\ Sd\left(D\right)=1.5\end{array}$

Explanation of Solution

Calculation:

Using the given information as stated in sub part a

  $\begin{array}{l}T~N\left(7,{0.9}^2\right)\\ C~N\left(10,{1.2}^2\right)\end{array}$

Both T and C are independent, so

  $\begin{array}{l}E\left[D\right]=E\left[C-T\right]=E\left[C\right]-E\left[T\right]\\ E\left[D\right]=10-7=3\\ V\left[D\right]=V\left[C-T\right]=V\left[C\right]+V\left[T\right]\\ V\left[D\right]={1.2}^2+{0.9}^2=2.25\end{array}$

So,

  $D~N\left(3,{1.5}^2\right)$

(c)

To determine

To find: Probability that a cap will break during fastening.

Answer to Problem 4CRE

The probability is 0.2275.

Explanation of Solution

Calculation:

From sub part b

  $D~N\left(3,{1.5}^2\right)$

The cap breaks while fastening means that the strength of the cap is less than that of the capping machine torque, so the probability can be calculated as follows

  $\begin{array}{l}P\left(C<T\right)=P\left(C-T<0\right)\\ P\left(D<0\right)\end{array}$

Finding the probability

  $\begin{array}{l}P\left(D<0\right)=P\left(z<\frac{0-3}{1.5}\right)\\ =P\left(z<-2\right)\\ =P\left(z>2\right)\\ =1-P\left(z<2\right)\end{array}$

Reading from z tables

  $\begin{array}{l}P\left(z<2\right)=0.97725\\ P\left(z>2\right)=1-0.97725\\ P\left(z>2\right)=0.02275\end{array}$

Thus, the required probability is 0.02275.