# Solved – 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? Contents

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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# Solved – 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? 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):
``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.