On an exam, the mean score is 78 points, with a standard deviation of 6 points. Assuming normal distribution of the scores, approximately what percentage of students received more than 85?A.12% B.17% C.8% D.9% E.None of the above

To find the percentage of students who scored more than 85, we need to convert the score of 85 to a z-score using the formula:

\[ z = \frac{{X - \mu}}{{\sigma}} \]

Where: - \( X \) is the score we want to convert (85 in this case), - \( \mu \) is the mean of the scores (78), - \( \sigma \) is the standard deviation (6).

Plugging the values into the formula:

\[ z = \frac{{85 - 78}}{{6}} = \frac{7}{6} \approx 1.17 \]

Now, we refer to a standard normal distribution table (z-table) or use a calculator to find the proportion of the area under the curve to the right of \( z = 1.17 \). In most z-tables, the value given is for the area to the left of the z-score. We need to subtract this value from 1 to find the area to the right.

Looking up 1.17 on the z-table, we find that the area to the left is approximately 0.8790 (or 87.90%). To find the percentage of students who scored above 85, we subtract this from 1 (or 100%).

\[ 1 - 0.8790 = 0.1210 \]

Convert this to a percentage:

\[ 0.1210 \times 100\% = 12.1\% \]

So, approximately 12.1% of students scored more than 85 points.

The closest answer from the options is A.12%.