Answer:
84%.
Step-by-step explanation:
We have been given that a A class's exam scores are normally distributed. The average score is 65 and the standard deviation is 6.
We will use the z-score formula to find the z-score corresponding to raw score of 71.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Raw score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
Upon substituting our given values in z-score formula we will get,
[tex]z=\frac{71-65}{6}[/tex]
[tex]z=\frac{6}{6}=1[/tex]
Since we know that 68-95-99.7 rule states that approximately 68%, 95% and 99.7% of data lies within one, two and three standard deviation of mean respectively.
Since 68% of data lies within one standard deviation of mean. Now we subtract 68% from 100% and divide the result by 2.
[tex]\frac{100\%-68\%}{2}=\frac{32\%}{2}=16\%[/tex]
Now we will add 16% to 68% to get the percentage of students that scored below 71.
[tex]\text{The percentage of students scored below 71}=68\%+16\%[/tex]
[tex]\text{The percentage of students scored below 71}=84\%[/tex]
Therefore, approximately 84% of the students scored below 71.
