At a furniture factory,tables must be assembled,finished,and packaged before they can be shipped to stores.Based on past experience,the manager finds that the means and standard deviations (in minutes)of the times for each phase are as shown in the table: $$\begin{array} { l | l | l }
\text { Phase } & \text { Mean } & \text { SD } \
\hline \text { Assembly } & 25.4 & 2.6 \
\text { Finishing } & 35.5 & 3.6 \
\text { Packaging } & 14.6 & 1.8
\end{array}$$ What are the mean and standard deviation of the total time to prepare a table for shipping? Assume that the times for each phase are independent.
A)? = 46.03 min,? = 4.79 min
B)? = 75.5 min,? = 22.96 min
C)? = 75.5 min,? = 4.79 min
D)? = 46.03 min,? = 8 min
E)? = 75.5 min,? = 8 min

Question

At a furniture factory,tables must be assembled,finished,and packaged before they can be shipped to stores.Based on past experience,the manager finds that the means and standard deviations (in minutes)of the times for each phase are as shown in the table: $$\begin{array} { l | l | l } 
\text { Phase } & \text { Mean } & \text { SD } \
\hline \text { Assembly } & 25.4 & 2.6 \
\text { Finishing } & 35.5 & 3.6 \
\text { Packaging } & 14.6 & 1.8
\end{array}$$ What are the mean and standard deviation of the total time to prepare a table for shipping? Assume that the times for each phase are independent.
A)? = 46.03 min,? = 4.79 min
B)? = 75.5 min,? = 22.96 min
C)? = 75.5 min,? = 4.79 min
D)? = 46.03 min,? = 8 min
E)? = 75.5 min,? = 8 min

Quizplus · Accepted Answer

The mean total time is the sum of the individual means: 25.4 + 35.5 + 14.6 = 75.5 minutes. The standard deviation of the total time, assuming independence, is the square root of the sum of the squares of the individual standard deviations: sqrt((2.6)^2 + (3.6)^2 + (1.8)^2) = sqrt(6.76 + 12.96 + 3.24) = sqrt(22.96) = 4.79 minutes.

At a Furniture Factory,tables Must Be Assembled,finished,and Packaged Before They