In a statistics class, the mean and standard deviation of the scores obtained by students on a test were found to be 55 and 12 , respectively. If the scores are normally distributed, what percentage of students scored below 60 ?Option: 1 $22.36 \%$Option: 2 $66.28 \%$Option: 3 $78.65 \%$Option: 4 $\quad 78.12 \%$

To determine the percentage of students who scored below 60 , we need to calculate the z-score corresponding to that value and then find the area under the normal curve to the left of that z-score.

The z-score is calculated using the formula:

$\mathrm{ z=(x-\mu) / \sigma }$
where:

$x$ is the value we want to find the percentage below (in this case, 60),
$\mu$ is the mean (55), and
$\sigma$ is the standard deviation (12).

Plugging in the values:

\mathrm{ \begin{aligned} & z=(60-55) / 12 \\ & z=5 / 12 \\ & z \approx 0.4167 \end{aligned} }

Now we need to find the area to the left of this z-score in the standard normal distribution table (also known as the z-table) or by using a statistical software. The area represents the percentage of students who scored below 60 .

Looking up the z-score of 0.4167 in the z-table, we find that the area to the left is approximately 0.6628 .

So, approximately 66.28% of the students scored below 60 on the test.