Chapter 6, Problem R6.4RE

(a)

To determine

Probability for the randomly selected cap has strength greater than 11 inch − pounds.

Answer to Problem R6.4RE

Probability that for the randomly selected cap has strength greater than 11 inch − pounds is 0.2033.

Explanation of Solution

Given information:

For Cap strength C :

Cap strength, $x=11\text{inch}-\text{pounds}$

Mean, $\mu =10\text{inch}-\text{pounds}$

Standard deviation, $\sigma =1.2\text{inch}-\text{pounds}$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{11-10}{1.2}=0.83$

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

See the row that starts with 0.8 and the column that starts with .03 of the standard normal probability table for $P\left(z<0.83\right)$ .

  $\begin{array}{l}P\left(x>11\right)=P\left(z<0.83\right)\\ =1-P\left(z>0.83\right)\\ =1-0.7967\\ =0.2033\\ =20.33\%\end{array}$

Thus,

There are 20.33% chances that the randomly selected cap has strength greater than 11 inch − pounds.

(b)

To determine

Whether it is reasonable to assume the cap strength and torque applied by the machine are independent.

Answer to Problem R6.4RE

Yes, because both cap strength and torque are applied by different machines.

Explanation of Solution

Given information:

T : capping − machine torque

C : cap strength

Calculations:

We know that

T represents the capping − machine torque and C represents the cap strength.

We also know

Both T and C follow Normal distribution.

Also,

Both the torque applied and the strength of the caps vary.

Since the cap is applied by a different machine than the torque.

Thus,

It is reasonable is reasonable to assume the cap strength and torque applied by the machine are independent.

(c)

To determine

Mean and standard deviation of the random variable D .

Answer to Problem R6.4RE

For random variable D :

Mean,

  ${\mu }_D=3\text{inch}-\text{pound}$

Standard deviation,

  ${\sigma }_D=1.5\text{inch}-\text{pound}$

Explanation of Solution

Given information:

T : capping − machine torque

Such that

Mean,

  ${\mu }_T=7\text{inch}-\text{pound}$

Standard deviation,

  ${\sigma }_T=0.9\text{inch}-\text{pound}$

C : cap strength

Such that

Mean,

  ${\mu }_C=10\text{inch}-\text{pound}$

Standard deviation,

  ${\sigma }_C=1.2\text{inch}-\text{pound}$

Calculations:

For independent variables X and Y ,

Property mean:

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

Property variance:

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

For random variable,

We have

  $D=C-T$

Mean of the variable D ,

  ${\mu }_D={\mu }_{C-T}={\mu }_C-{\mu }_T=10-7=3\text{inch}-\text{pounds}$

Variance of the variable D ,

  ${\sigma }_D^2={\sigma }_{C-T}^2={\sigma }_C^2+{\sigma }_T^2={\left(1.2\right)}^2+{\left(0.9\right)}^2=2.25\text{inch}-\text{pounds}$

We also know

The standard deviation is the square root of the variance:

  ${\sigma }_D=\sqrt{{\sigma }_D^2}=\sqrt{2.25}=1.5\text{inch}-\text{pounds}$

(d)

To determine

Probability for the randomly selected cap will break while being fastened by the machine.

Answer to Problem R6.4RE

Probability that the randomly selected cap will break while being fastened by the machine is 0.0228.

Explanation of Solution

Given information:

From Part (c),

For random variable D :

Mean,

  ${\mu }_D=3\text{inch}-\text{pound}$

Standard deviation,

  ${\sigma }_D=1.5\text{inch}-\text{pound}$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{0-3}{1.5}=-2.00$

Use table A, to find the corresponding probability.

  $P\left(T>C\right)=P\left(C-T<0\right)=P\left(z<-2.00\right)=0.0228=2.28\%$

Thus,

There are 2.28% chances for the randomly selected cap will break while being fastened by the machine and the probability is 0.0228.