91Ó°ÊÓ

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.

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

Here,

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

Therefore, the mean of $s$

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

Let's compute the standard deviation of s

Here,

$$\begin{gathered} {\mathrm{σ}}_Y=2 \\ {\mathrm{σ}}_Y=4 \end{gathered}$$

Standard deviation of S.

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

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.

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

So, $P(X+Y≤30)$

First we standardize $S=30$

$$\begin{gathered} z=\frac{x-{\mathrm{μ}}_s}{{\mathrm{σ}}_s} \\ =\frac{30−31}{4.47} \\ =-0.22 \end{gathered}$$

Using a table of typical normal probabilities as a guide,

$P(S≤30)$

$$\begin{gathered} =P(z≤-0.22) \\ =P(zâ‰\yen 0.22) \\ =1−P(z<0.22) \\ =1−0.5871 \\ =0.4129 \end{gathered}$$