A time-and-motion study measures the time required for an assembly-line worker to perform a repetitive task. The data show that the time $X$required to bring a part from a bin to its position on an automobile chassis follows a Normal distribution with mean $11$seconds and standard deviation $2$seconds. The time $Y$required to attach the part to the chassis follows a Normal distribution with mean $20$seconds and standard deviation $4$seconds. The study finds that the times required for the two steps are independent.

a. Describe the distribution of the total time required for the entire operation of positioning and attaching a randomly selected part.

b. Management’s goal is for the entire process to take less than $30$seconds. Find the probability that this goal will be met for a randomly selected part.

Step by step solution

01

Part(a) Step 1 : Given Information 

Given :

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

$Y$: Time required attaching the part to the chassis

Mean,

${\mu }_X=11$

role="math" localid="1654242465628" ${\mu }_Y=20$

Standard deviation,

${\sigma }_X=2$

${\sigma }_Y=4$

02

Part(a) Step 2 : Simplification  

If $X\mathrm{and}Y$are independent, Property mean:

role="math" localid="1654242716030" ${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

Property variance:

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

Mean of $X+Y$,

$$\begin{gathered} {\mu }_{X+Y}={\mu }_X+{\mu }_Y \\ =11+20 \\ =31\mathrm{seconds} \end{gathered}$$

Standard deviation of $X+Y$,

$$\begin{gathered} {\sigma }_{x+y}=\sqrt{{{\sigma }^2}_X+{{\sigma }^2}_Y} \\ =\sqrt{{\left(2\right)}^2+{\left(4\right)}^2} \\ \approx \mathbf{4}\mathbf{.}\mathbf{4721}\mathrm{seconds} \end{gathered}$$

03

Part(b) Step 1 : Given Information 

Given :

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

$Y$: Time required attaching the part to the chassis

Mean,

${\mu }_X=11$

${\mu }_Y=20$

Standard deviation,

${\sigma }_X=2$

${\sigma }_Y=4$

04

Part(b) Step 2 : Simplification  

The distribution is a normal distribution, as we know.

Calculate the $z-score$, which is:

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

Table $-A$: Find the corresponding value.

$$\begin{gathered} P(X+Y<30)=P(Z<-0.22) \\ =\mathbf{}\mathbf{0}\mathbf{.}\mathbf{4129}\mathbf{} \end{gathered}$$

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