Chapter 6.2, Problem 65E

(a)

To determine

Discuss the distribution of the total time required for the entire operation.

Answer to Problem 65E

The distribution of the total time required is Normal distribution.

Explanation of Solution

Given information:

X : Time required bringing a part from a bin to its position

Y : Time required attaching the part to the chassis

For X :

Mean,

  ${\mu }_X=11$

Standard deviation,

  ${\sigma }_X=2$

For Y :

Mean,

  ${\mu }_Y=20$

Standard deviation,

  ${\sigma }_Y=4$

If X and Y are independent,

Property mean:

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

Property variance:

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

Thus,

We have

Mean of X + Y ,

  ${\mu }_{X+Y}={\mu }_X+{\mu }_Y=11+20=31\text{seconds}$

Standard deviation of X + Y ,

  ${\sigma }_{X+Y}=\sqrt{{\sigma }_X^2+{\sigma }_Y^2}=\sqrt{{\left(2\right)}^2+{\left(4\right)}^2}=\sqrt{20}\approx 4.4721\text{seconds}$

Thus,

The distribution will be the Normal distribution.

(b)

To determine

Probability for management’s goal will be met for a randomly selected part.

Answer to Problem 65E

Probability that the management’s goal will be met for a randomly selected part is 0.4129.

Explanation of Solution

Given information:

X : Time required bringing a part from a bin to its position

Y : Time required attaching the part to the chassis

For X :

Mean,

  ${\mu }_X=11$

Standard deviation,

  ${\sigma }_X=2$

For Y :

Mean,

  ${\mu }_Y=20$

Standard deviation,

  ${\sigma }_Y=4$

Time taken,

  $x=30$

Calculations:

If X and Y are independent,

Property mean:

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

Property variance:

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

Thus,

We have

Mean of X + Y ,

  ${\mu }_{X+Y}={\mu }_X+{\mu }_Y=11+20=31\text{seconds}$

Standard deviation of X + Y ,

  ${\sigma }_{X+Y}=\sqrt{{\sigma }_X^2+{\sigma }_Y^2}=\sqrt{{\left(2\right)}^2+{\left(4\right)}^2}=\sqrt{20}\approx 4.4721\text{seconds}$

We know that

The distribution is a normal distribution.

Now,

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{30-31}{4.4721}\approx -0.22$

Find the corresponding value from Table − A:

  $P\left(X+Y<30\right)=P\left(z<-0.22\right)=0.4129$

Thus,

Probability that the management’s goal will be met for a randomly selected part is 0.4129.