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?

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#### Best Answer

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**):

$$ boldsymbol{s} = (s_1, ldots, s_n), quadmathrm{ans} = frac{#left{s_icolon s_i > left( bar{boldsymbol{s}} + sqrt{frac{1}{n-1} (boldsymbol{s} – bar{boldsymbol{s}})' (boldsymbol{s} – bar{boldsymbol{s}}}) right)right}}{n} cdot 100% $$ 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.

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