Provide an appropriate response.
-The manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one Monday. The frequency distribution below summarizes the results. Find
The standard deviation. Round your answer to one decimal place. $\begin{array}{c|c}
\begin{array}{r}
\text { Waiting time } \
\text { (minutes) }
\end{array} & \begin{array}{c}
\text { Number of } \
\text { customer }
\end{array} \
\hline 0 \text {-under } 4 & 13 \
4 \text {-under } 8 & 12 \
8 \text {-under } 12 & 12 \
12 \text {-under } 16 & 12 \
16 \text {-under } 20 & 0 \
20 \text {-under } 24 & 1
\end{array}$
A)$4.7$
B) $7.0$
C) $4.9$
D) $5.2$

Question

Provide an appropriate response.
-The manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one Monday. The frequency distribution below summarizes the results. Find
The standard deviation. Round your answer to one decimal place. $\begin{array}{c|c}
\begin{array}{r}
\text { Waiting time } \
\text { (minutes) }
\end{array} & \begin{array}{c}
\text { Number of } \
\text { customer }
\end{array} \
\hline 0 \text {-under } 4 & 13 \
4 \text {-under } 8 & 12 \
8 \text {-under } 12 & 12 \
12 \text {-under } 16 & 12 \
16 \text {-under } 20 & 0 \
20 \text {-under } 24 & 1
\end{array}$  
A)$4.7$
B) $7.0$
C) $4.9$
D) $5.2$

Quizplus · Accepted Answer

To calculate the standard deviation for grouped data, we first need to find the mean (average) waiting time, then use it to calculate the variance, and finally take the square root of the variance to find the standard deviation. The midpoint of each class is used to represent the waiting times for that class. For example, the midpoint for the first class (0-under 4 minutes) is (0+4)/2 = 2 minutes. The process involves multiplying each midpoint by the number of customers in that class, summing these products, dividing by the total number of customers to find the mean, then using the mean to calculate the variance and standard deviation. The correct standard deviation, rounded to one decimal place, is approximately 4.9 minutes, indicating the average amount by which the waiting times deviate from the mean waiting time.

Provide an Appropriate Response A) 4.74.74.7 B) 7.07.07.0 C) 4.94.94.9 D) 5.25.25.2

Provide an Appropriate Response A) $4.7$ B) $7.0$ C) $4.9$ D) $5.2$