# What percentage of the students scored more than one standard deviation above the mean?

I was given a question for an assignment but I don't understand whether or not I have the right answer...

The question is this-

What percentage of the students scored more than one standard deviation above the mean?

I was given a data set of 50 scores of students in a statistics course and calculated the following using minitab.

• Mean- 62.46
• Median- 61.00
• Variance 237.76

The standard deviation calculated was 5.7035 as I took the square root of the variance.

The score at one standard deviation above the mean would be 68.1635

Is my answer supposed to be 15.8%? As when looking at a symmetrical distribution curve we can see that one standard deviation is 34.1% so I took the next three percentages and added them to find the percent

13.6 + 2.1 + 0.1 = 15.8%

Or am I suppose to use 68.1635 to figure out the percentage? • I am sorry, the variance is 237 and its square root is 5.70? Something's not right there. – JohnK Jan 24 '16 at 17:03

Why are you using the normality assumption? You do not know the distribution of scores in the sample. So, given a dataset (let us denote it with s, a vector of the student scores), the following routine will give you the exact result for any distribution (below is the implementation in R):
$$\boldsymbol{s} = (s_1, \ldots, s_n), \quad\mathrm{ans} = \frac{\#\left\{s_i\colon s_i > \left( \bar{\boldsymbol{s}} + \sqrt{\frac{1}{n-1} (\boldsymbol{s} - \bar{\boldsymbol{s}})' (\boldsymbol{s} - \bar{\boldsymbol{s}}}) \right)\right\}}{n} \cdot 100\%$$ where $\bar{\boldsymbol{s}} = \frac{1}{n} \sum s_i$ is the arithmetic mean and $\#\{\cdot\}$ just counts the elements of a set that satisfy the condition.
sum(s > mean(s) + sd(s)) / length(s) * 100

This thing does exactly what it says on the tin: s > mean(s) + sd(s) returns TRUE for those guys who were above one SD, sum counts them (TRUE is converted to 1 and FALSE to 0), and then you compute the percentage.