Question

Time and motion 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 required to bring a part from a bin to its position on an automobile chassis varies from car to car according to a Normal distribution with mean $11$seconds and standard deviation $2$seconds. The time 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 part that takes a long time to position, for example, does not take more or less time to attach than other parts.

(a) Find the mean and standard deviation of the time required for the entire operation of positioning and attaching the 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. Show your work

Short answer

(a) Mean and Standard deviation are ${\mu }_S=31and{\sigma }_s=4.47$

(b) The probability that the entire process will take less than$30$seconds is$0.4129$

Step-by-step solution

Part (a) Step 1: Given Information 

A time and motion study measures the time required for an assembly line worker to perform a repetitive task.

Given time required to bring a part from a bin to its position on an automobile chassis X follows a Normal distribution with mean $11$seconds and standard deviation $2$seconds.

The time required to attach the part to the chassis $Y$follows a Normal distribution with mean $20$seconds and standard deviation $4$seconds.

We have to find that the mean and standard deviation of the time required for the entire operation of positioning and attaching the part.

Part (a) Step 2: Calculate the mean and standard deviation 

Consider$S=X+Y$asa random variable which shows the time required for the entire operation of positioning and attaching the part.

Here,

$$\begin{gathered} {\mu }_X=11 \\ {\mu }_Y=20 \end{gathered}$$

Therefore, the mean of $s$

$$\begin{gathered} {\mu }_S={\mu }_X+{\mu }_Y \\ =11+20 \\ =31 \end{gathered}$$

Let's compute the standard deviation of s

Here,

$$\begin{gathered} {\sigma }_Y=2 \\ {\sigma }_Y=4 \end{gathered}$$

Standard deviation of S.

$$\begin{gathered} {\sigma }_S=\sqrt{{{\sigma }_X}^2+{{\sigma }_Y}^2} \\ =\sqrt{2^2+4^2} \\ =4.47 \end{gathered}$$

Part (b) Step 1: Given  Information

Given in the question that Management’s goal is for the entire process to take less than $30$seconds.

We have to find the probability that this goal will be met.

Part (b) Step 2: Explanation 

Let's find the probability that the entire process will take less than 30 seconds.

So, $P(X+Y\le 30)$

First we standardize $S=30$

$$\begin{gathered} z=\frac{x-{\mu }_s}{{\sigma }_s} \\ =\frac{30-31}{4.47} \\ =-0.22 \end{gathered}$$

Using a table of typical normal probabilities as a guide,

$P(S\le 30)$

$$\begin{gathered} =P(z\le -0.22) \\ =P(z\ge 0.22) \\ =1-P(z<0.22) \\ =1-0.5871 \\ =0.4129 \end{gathered}$$