Applying torque A machine fastens plastic screw-on caps onto containers of motor oil. If the machine applies more torque than the cap can withstand, the cap will break. Both the torque applied and the strength of the caps vary. The capping-machine torque T follows a Normal distribution with mean 7 inch-pounds and standard deviation 0.9 inchpound. The cap strength C (the torque that would break the cap) follows a Normal distribution with mean 10 inch-pounds and standard deviation 1.2 inch-pounds.

a. Find the probability that a randomly selected cap has a strength greater than 11 inchpounds.

b. Explain why it is reasonable to assume that the cap strength and the torque applied by the machine are independent.

c. Let the random variable D=C-T. Find its mean and standard deviation.

d. What is the probability that a randomly selected cap will break while being fastened by the machine?

For Cap strength C:

Cap strength, $x=11$inch pounds

Mean, $\mu =10$inch - pounds

Standard deviation, $\sigma =1.2$inch pounds

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 $P(z<0.83)$.

role="math" localid="1654198722700" $$\begin{gathered} P(x>11)\&=P(z<0.83)\backslash \backslash \\ =1-P(z>0.83)\backslash \backslash \\ =1-0.7967\backslash \backslash \\ =0.2033\backslash \backslash \\ =20.33\backslash \% \end{gathered}$$

Thus,

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

T : capping - machine torque

C: cap strength

We know that

T represents the capping - machine torque and $\mathrm{C}$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.

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,

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

For independent variables X and Y,

Property mean:

${\mu }_{ux+b\gamma }=a\mu x+b\mu \gamma$

Property variance:

$a_{uX+bY}^2=a^2{\sigma }_X^2+b^2{\sigma }_Y^2$

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,

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

We also know

The standard deviation is the square root of the variance:

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

From Part (c),

For random variable D :

Mean,

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

Standard deviation,

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

Calculate the z- score,

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

Use table A, to find the corresponding probability.

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

Thus,

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