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Given in the question that, A machine fastens plastic screwon 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 $0.9$inch-pounds. 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.

we need to explain that why it is reasonable to assume that the cap strength and the torque applied by the machine are independent.

$T-$The capping machine torque

$\mathrm{C}$-The cap strength

Both of them follow normal distribution

Mean of $\mathrm{T}=7$

Standard deviation of$\mathrm{T}=0.9$

Mean of $C=10$

Standard deviation of$C=1.2$

When two events are independent

$E[A鈭�B]=E\left[A\right]鈭�E\left[B\right]$

A machine fastens plastic screwon 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$0.9$inch-pounds. 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.

We need to find the mean and standard deviation of random variable$D=C鈭�T.$

Using the given information as stated in sub part a

$T~N\left(7,{0.9}^2\right)$

$C~N\left(10,{1.2}^2\right)$

Both $\mathrm{T}$and $\mathrm{C}$are independent, so

$E\left[D\right]=E[C鈭�T]=E\left[C\right]鈭�E\left[T\right]$

$E\left[D\right]=10鈭�7=3$

$V\left[D\right]=V[C鈭�T]=V\left[C\right]+V\left[T\right]$

$V\left[D\right]={1.2}^2+{0.9}^2=2.25$

So,

$D~N\left(3,{1.5}^2\right)$

A machine fastens plastic screwon 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 $0.9$inch-pounds. 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.

From sub part b

$D~N\left(3,1.5^2\right)$

Because the cap's strength is less than the capping machine torque when it breaks while fastening, the likelihood can be computed as follows:

$P(C<T)=P(C鈭�T<0)$

$P(D<0)$

Finding the probability

$P(D<0)=P(z<\frac{0-3}{1.5})$

localid="1649604622419" $=P(z<-2)$

localid="1649604614004" $=P(z>2)$

$=1-P(z<2)$

Reading from z tables

$P(z<2)=0.97725$

$P(z<2)=0.97725$

$P(z<2)=0.97725$