Question
During clinical trials of a new drug intended to reduce the risk of heart attack, the following data indicate the occurrence of adverse reactions among 1 comma 200 adult male trial members. What is the probability that an adult male using the drug will experience nausea?
Start 7 By 2 Table 1st Row 1st Column Adverse Reaction 2nd Column Number 2nd Row 1st Column Heartburn 2nd Column 18 3rd Row 1st Column Headache 2nd Column 13 4st Row 1st Column Dizziness 2nd Column 8 5st Row 1st Column Urinary problems 2nd Column 5 6st Row 1st Column Nausea 2nd Column 26 7st Row 1st Column Abdominal pain 2nd Column 18 EndTable
Start 7 By 2 Table 1st Row 1st Column Adverse Reaction 2nd Column Number 2nd Row 1st Column Heartburn 2nd Column 18 3rd Row 1st Column Headache 2nd Column 13 4st Row 1st Column Dizziness 2nd Column 8 5st Row 1st Column Urinary problems 2nd Column 5 6st Row 1st Column Nausea 2nd Column 26 7st Row 1st Column Abdominal pain 2nd Column 18 EndTable
Answers
GPT-4o mini
To find the probability that an adult male using the drug will experience nausea, we can use the following formula:
\[
P(\text{Nausea}) = \frac{\text{Number of cases of nausea}}{\text{Total number of trial members}}
\]
From the data provided:
- Number of cases of nausea = 26
- Total number of trial members = 1200
Now, we can compute the probability:
\[
P(\text{Nausea}) = \frac{26}{1200}
\]
We can simplify this fraction:
\[
P(\text{Nausea}) = \frac{26}{1200} = \frac{13}{600} \approx 0.02167
\]
Therefore, the probability that an adult male using the drug will experience nausea is approximately **0.02167** or **2.17%**.
\[
P(\text{Nausea}) = \frac{\text{Number of cases of nausea}}{\text{Total number of trial members}}
\]
From the data provided:
- Number of cases of nausea = 26
- Total number of trial members = 1200
Now, we can compute the probability:
\[
P(\text{Nausea}) = \frac{26}{1200}
\]
We can simplify this fraction:
\[
P(\text{Nausea}) = \frac{26}{1200} = \frac{13}{600} \approx 0.02167
\]
Therefore, the probability that an adult male using the drug will experience nausea is approximately **0.02167** or **2.17%**.
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