Question

In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been properly drilled in the boxes and in ten minutes if the holes must be redrilled. Twenty gearboxes are in stock, 2 with improperly drilled holes. Five gearboxes must be selected from the 20 that are available for installationcin the next five robots.

(a) Find the probability that all 5 gearboxes will fit properly.

(b) Find the mean, variance, and standard deviation of the time it takes to install these 5 gearboxes.

Answer 1

Solution

(a) Find the probability that all 5 gearboxes will fit properly.

Let X be the number of gearboxes fitting properly  \( X\sim Hp(5,18,20) \)

we have  \( P(X=x)=\frac{C_M^x\times C_{N-M}^{n-x}}{C_N^n}=\frac{C_{18}^x\times C_2^{5-x}}{C_{20}^5} \)

\( \implies P(X=5)=\frac{C_{18}^5\times \:C_2^0}{C_{20}^5}=\frac{18!\times 15!}{13!\times 20!}=\frac{14\times 15}{19\times \:20}=0.55 \)

Therefore.  \( P(X=5)=0.55 \)

(b) Find the mean, variance, and standard deviation of the time

X number of gearboxes fitting properly 

Then  \( T(X)=X\times 1+10\times(5-X)=50-9X \)

So, \( E(T(X))=n\times\frac{M}{N}\times(-9)+50=5\times\frac{18}{20}\times(-9)+50=9.5 \)

\( V(T(X))=V(50-9X)=81\times V(X) \)

\( V(X)=\frac{20-5}{20-1}\times 5\times\frac{18}{20}\times(1-\frac{18}{20})=0.35 \)

\( \implies V(T(X))=81\times 0.35=28.77 \iff \delta(T(X))=\sqrt{V(T(X))}=\sqrt{28.77}=5.36 \)Therefore.  \( E(T)=9.5\hspace{2mm},V(T)=28.77\hspace{2mm},\delta(T(X))=5.36 \)

Therefore. 

a). P(X=5)=0.55

b). \( E(T)=9.5\hspace{2mm},V(T)=28.77\hspace{2mm},\delta(T(X))=5.36 \)