Question

A class'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.

Answer 1

Standardising X=71 to find z-score


`z-score= (71-65)/(6) =1`

reading from z-table


`P(Z\ \textless \ 1)=0.8413=84.13%`

Students score below 71 is 84.13%

Answer 2

84%.

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.

`z=(x-\mu)/(\sigma)`, where,

`z=\text{z-score}`,

`x=\text{Raw score}`,

`\mu=\text{Mean}`,

`\sigma=\text{Standard deviation}`.

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

`z=(71-65)/(6)`

`z=(6)/(6)=1`

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.

`(100\%-68\%)/(2)=(32\%)/(2)=16\%`

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

`\text{The percentage of students scored below 71}=68\%+16\%`

`\text{The percentage of students scored below 71}=84\%`

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