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

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.

$z=\frac{x-\mu}{\sigma}$, 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,

$z=\frac{71-65}{6}$

$z=\frac{6}{6}=1$

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.

$\frac{100\%-68\%}{2}=\frac{32\%}{2}=16\%$

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.

Answer 2

Standardising X=71 to find z-score

$z-score= \frac{71-65}{6} =1$

reading from z-table

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

Students score below 71 is 84.13%