Find the midrange for the given sample data.
-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 mean waiting time.
Round your answer to one decimal place. $$\begin{array}{l}
\text { Waiting time Number of }\
\begin{array} { r c }
\text { (minutes) } & \text { customers } \
0 - 3 & 10 \
4 - 7 & 13 \
8 - 11 & 12 \
12 - 15 & 5 \
16 - 19 & 7 \
20 - 23 & 1 \
24 - 27 & 2
\end{array}
\end{array}$$
A) 9.4 min
B) 13.5 min
C) 7.1 min
D) 9.3 min

Question

Find the midrange for the given sample data.
-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 mean waiting time.
Round your answer to one decimal place. $$\begin{array}{l}
\text { Waiting time Number of }\
\begin{array} { r c } 
\text { (minutes) } & \text { customers } \
0 - 3 & 10 \
4 - 7 & 13 \
8 - 11 & 12 \
12 - 15 & 5 \
16 - 19 & 7 \
20 - 23 & 1 \
24 - 27 & 2
\end{array}
\end{array}$$
A) 9.4 min
B) 13.5 min
C) 7.1 min
D) 9.3 min

Quizplus · Accepted Answer

To find the midrange, we need to find the average of the highest and lowest values.
The lowest value is 0 and the highest value is 27. 
Midrange = (0 + 27) / 2 = 13.5
Therefore, the midrange is 13.5 minutes.

To find the mean waiting time, we need to use the formula:

Mean = (sum of (frequency × midpoint)) ÷ (sum of frequency)

Using this formula, we get:

Mean = ((10 × 1.5) + (13 × 5.5) + (12 × 9.5) + (5 × 13.5) + (7 × 17.5) + (1 × 21.5) + (2 × 25.5)) ÷ 50
Mean = 451 ÷ 50
Mean = 9.02

Rounding to one decimal place, we get Mean = 9.0 minutes.

Therefore, the best choice is D, 9.3 min, which is the closest rounded value to the calculated mean of 9.0 minutes.

Find the Midrange for the Given Sample Data A) 94 Min B) 13

Find the Midrange for the Given Sample Data A) 94 Min
B) 13