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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,

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

Therefore, the mean of $s$

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

Let's compute the standard deviation of s

Here,

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

Standard deviation of S.

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

Given in the question that Management鈥檚 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, $P(X+Y鈮�30)$

First we standardize $S=30$

$$\begin{gathered} z=\frac{x-{渭}_s}{{蟽}_s} \\ =\frac{30鈭�31}{4.47} \\ =-0.22 \end{gathered}$$

Using a table of typical normal probabilities as a guide,

$P(S鈮�30)$

$$\begin{gathered} =P(z鈮�-0.22) \\ =P(z鈮�0.22) \\ =1鈭�P(z<0.22) \\ =1鈭�0.5871 \\ =0.4129 \end{gathered}$$