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-four gearboxes are in stock, 6 with improperly drilled holes. Five gearboxes must be selected from the 24 that are available for installation in the next five robots. (Round your answers to four decimal places.) (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

Answer:

The right answer is:

(a) 0.1456

(b) 18.125, 69.1202, 8.3139

Step-by-step explanation:

Given:

N = 24

n = 5

r = 7

The improperly drilled gearboxes "X".

then,

⇒ $P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}$

(a)

P (all gearboxes fit properly) = $P(x=0)$

                                               = $\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}$

                                               = $0.1456$

(b)

According to the question,

$X = 91+5$

Mean will be:

⇒ $\mu = E(x)$

       $=E(91+5)$

       $=9E(1)+5$

       $=9.\frac{nr}{N}+5$

       $=9.\frac{5.7}{24} +5$

       $=18.125$

Variance will be:

⇒ $\sigma^2=Var(X)$

         $=V(9Y+5)$

         $=81.V(Y)$

         $=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}$

         $=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}$

         $=69.1202$

Standard deviation will be:

⇒ $\sigma = \sqrt{69.1202}$

       $=8.3139$