Provide an appropriate response.
-On a recent day during the flu season, the Midtown Medical Clinic saw many patients. Patients had their temperatures (in °F) taken. The distribution of temperatures is given below. Determine the mean temperature of a sample of 100 patients. $\begin{array}{|c|c|}
\hline \text { Temperatures } & \text { Frequency } \
\hline 95.6-96.49 & 1 \
\hline 96.5-97.39 & 3 \
\hline 97.4-98.29 & 19 \
\hline 98.3-99.19 & 28 \
\hline 99.2-100.09 & 35 \
\hline 100.1-100.99 & 12 \
\hline 101.0-101.89 & 2 \
\hline
\end{array}$

A) $99.08 ^ { \circ } \mathrm { F }$
B) $99.52 ^ { \circ } \mathrm { F }$
C) $98.75 ^ { \circ } \mathrm { F }$
D) $98.63 ^ { \circ } \mathrm { F }$

Question

Provide an appropriate response.
-On a recent day during the flu season, the Midtown Medical Clinic saw many patients. Patients had their temperatures (in °F) taken. The distribution of temperatures is given below. Determine the mean temperature of a sample of 100 patients. $\begin{array}{|c|c|}
\hline \text { Temperatures } & \text { Frequency } \
\hline 95.6-96.49 & 1 \
\hline 96.5-97.39 & 3 \
\hline 97.4-98.29 & 19 \
\hline 98.3-99.19 & 28 \
\hline 99.2-100.09 & 35 \
\hline 100.1-100.99 & 12 \
\hline 101.0-101.89 & 2 \
\hline
\end{array}$

A) $99.08 ^ { \circ } \mathrm { F }$
B) $99.52 ^ { \circ } \mathrm { F }$
C) $98.75 ^ { \circ } \mathrm { F }$
D) $98.63 ^ { \circ } \mathrm { F }$

Quizplus · Accepted Answer

To find the mean temperature, calculate the midpoint of each temperature range, multiply each midpoint by the frequency of the range, sum these products, and then divide by the total number of observations (100). The calculation is as follows:- $95.6-96.49$ midpoint: $96.045$, product: $96.045 \times 1 = 96.045$- $96.5-97.39$ midpoint: $96.945$, product: $96.945 \times 3 = 290.835$- $97.4-98.29$ midpoint: $97.845$, product: $97.845 \times 19 = 1860.055$- $98.3-99.19$ midpoint: $98.745$, product: $98.745 \times 28 = 2764.86$- $99.2-100.09$ midpoint: $99.645$, product: $99.645 \times 35 = 3487.575$- $100.1-100.99$ midpoint: $100.545$, product: $100.545 \times 12 = 1206.54$- $101.0-101.89$ midpoint: $101.445$, product: $101.445 \times 2 = 202.89$Sum of products: $96.045 + 290.835 + 1860.055 + 2764.86 + 3487.575 + 1206.54 + 202.89 = 9909.8$Mean temperature: $9909.8 / 100 = 99.098 ^ { \circ } \mathrm { F }$Rounded to two decimal places, the mean temperature is $99.08 ^ { \circ } \mathrm { F }$.

Provide an Appropriate Response A) 99.08∘F99.08 ^ { \circ } \mathrm { F }99.08∘F

Provide an Appropriate Response A) $99.08 ^ { \circ } \mathrm { F }$