Two hundred simulation runs were completed using the probability of a machine breakdown from the table below. The average number of breakdowns from the simulation trials was 1.93 with a standard deviation of 0.20.
$$\begin{array} { | c | c | c | }
\hline \begin{array} { c }
\text { No. of } \
\text { breakdowns } \
\text { per week }
\end{array} & \text { Probability } & \begin{array} { c }
\text { Cumulative } \
\text { probability }
\end{array} \
\hline 0 & .10 & .10 \
\hline 1 & .25 & .35 \
\hline 2 & .36 & .71 \
\hline 3 & .22 & .93 \
\hline 4 & .07 & 1.00 \
\hline
\end{array}$$
-Construct a 95% confidence interval for the average number of machine breakdowns.
A) 1.896 to 1.938
B) 1.902 to 1.958.
C).1.877 to 1.943
D) 1.907 to 1.953.

Question

Two hundred simulation runs were completed using the probability of a machine breakdown from the table below. The average number of breakdowns from the simulation trials was 1.93 with a standard deviation of 0.20.
 $$\begin{array} { | c | c | c | } 
\hline \begin{array} { c } 
\text { No. of } \
\text { breakdowns } \
\text { per week }
\end{array} & \text { Probability } & \begin{array} { c } 
\text { Cumulative } \
\text { probability }
\end{array} \
\hline 0 & .10 & .10 \
\hline 1 & .25 & .35 \
\hline 2 & .36 & .71 \
\hline 3 & .22 & .93 \
\hline 4 & .07 & 1.00 \
\hline
\end{array}$$
-Construct a 95% confidence interval for the average number of machine breakdowns.
A) 1.896 to 1.938
B) 1.902 to 1.958.
C).1.877 to 1.943
D) 1.907 to 1.953.

Quizplus · Accepted Answer

The 95% confidence interval is calculated using the formula: mean ± (Z * (standard deviation / sqrt(n))). With a mean of 1.93, standard deviation of 0.20, and n = 200, the interval is 1.902 to 1.958.

Two Hundred Simulation Runs Were Completed Using the Probability of a Machine