91Ó°ÊÓ

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,

${\mathrm{μ}}_X=11$

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

Standard deviation,

${\mathrm{σ}}_X=2$

${\mathrm{σ}}_Y=4$

02

Part(a) Step 2 : Simplification  

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

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

Property variance:

${{\mathrm{σ}}^2}_{aX+bY}=a^2{\mathrm{μ}}_X+b^2{\mathrm{μ}}_Y$

Mean of $X+Y$,

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

Standard deviation of $X+Y$,

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

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,

${\mathrm{μ}}_X=11$

${\mathrm{μ}}_Y=20$

Standard deviation,

${\mathrm{σ}}_X=2$

${\mathrm{σ}}_Y=4$

04

Part(b) Step 2 : Simplification  

The distribution is a normal distribution, as we know.

Calculate the $z-score$, which is:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{30-31}{4.4721} \\ ≈-0.22 \end{gathered}$$

Table $-A$: Find the corresponding value.

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

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!