Questions pertain to the following table (Table 3):
$\begin{array}{c}
\text { Table } 3\
\text { Characteristics of Chemotherapy Patients }(N=100)\
\begin{array}{lcc}
\text { Characteristic } & M&(S D) & M d n \
\text { Age (years) } & 48.9&(9.8) & 47.0 \
\text { Body mass index (BM) }\left(\mathrm{kg} / \mathrm{m}^{2}\right) & 27.0&(6.0) & 25.1 \
\text { Number of positive nodes } & 3.4&(2.9) & 2.0 \
\text { Dose of cyclophosphamide }(\mathrm{mg}) & 1063.0&(477.0) & 1250.0 \
\text { Dose of doxorubicin (mg) } & 125.0&(53.0) & 125.0 \
\text { Degree of nausea, 0-100 scale } & 52.1&(25.0) & 52.0
\end{array}\end{array}$
-Refer to Table 3. Assume that the distribution for the variable degree of nausea is normally distributed. In such a case, out of the 100 sample members, approximately how many gave a nausea rating of 77 or higher?

Question

Questions pertain to the following table (Table 3):
 $\begin{array}{c}
\text { Table } 3\
\text { Characteristics of Chemotherapy Patients }(N=100)\
\begin{array}{lcc}
\text { Characteristic } & M&(S D) & M d n \
\text { Age (years) } & 48.9&(9.8) & 47.0 \
\text { Body mass index (BM) }\left(\mathrm{kg} / \mathrm{m}^{2}\right) & 27.0&(6.0) & 25.1 \
\text { Number of positive nodes } & 3.4&(2.9) & 2.0 \
\text { Dose of cyclophosphamide }(\mathrm{mg}) & 1063.0&(477.0) & 1250.0 \
\text { Dose of doxorubicin (mg) } & 125.0&(53.0) & 125.0 \
\text { Degree of nausea, 0-100 scale } & 52.1&(25.0) & 52.0
\end{array}\end{array}$
-Refer to Table 3. Assume that the distribution for the variable degree of nausea is normally distributed. In such a case, out of the 100 sample members, approximately how many gave a nausea rating of 77 or higher?

Quizplus · Accepted Answer

Since the distribution for the variable degree of nausea is assumed to be normal, we can use the mean and standard deviation to calculate the proportion of patients who gave a nausea rating of 77 or higher. 
First, we need to calculate the z-score: 
z = (77 - 52.1)/25 = 0.996 
Using a standard normal table or calculator, we can find that the proportion of patients who gave a nausea rating of 77 or higher is approximately 0.1587. 
So, the approximate number of patients who gave a nausea rating of 77 or higher is: 
0.1587 * 100 = 15.87, which rounds up to 16. 
Therefore, the best choice is C.

Questions Pertain to the Following Table (Table 3) -Refer to Table 3