Aclass's exam scores are normally distributed. if the average score is 65 and the standard deviation is 6, what percentage of students scored below 71? hint: use the 68-95-99.7 rule.

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.

, where,

,

,

,

.

Upon substituting our given values in z-score formula we will get,

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.

Now we will add 16% to 68% to get the percentage of students that scored below 71.

Therefore, approximately 84% of the students scored below 71.

table 3 represents a direct variation function. : )