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Two ordered data sets, each with 10 numbers. Find the sample standard deviation for each and for their sum.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "The following table gives the examination marks in {exam1} for a group of $n=\\var{n}$ students.
\n{exam1} | \n$\\var{r0[0]}$ | \n$\\var{r0[1]}$ | \n$\\var{r0[2]}$ | \n$\\var{r0[3]}$ | \n$\\var{r0[4]}$ | \n$\\var{r0[5]}$ | \n$\\var{r0[6]}$ | \n$\\var{r0[7]}$ | \n$\\var{r0[8]}$ | \n$\\var{r0[9]}$ | \nMean = $\\var{mean1}$ | \n
---|
Find the standard deviation for the {exam1} examination marks.
", "advice": "The solution to the first part is here – the other parts can be done in the same way.
\nFor {exam1} we have the mean is:
\n\\[\\simplify[]{({r0[0]} + {r0[1]} + {r0[2]} + {r0[3]} + {r0[4]} + {r0[5]} + {r0[6]} + {r0[7]} + {r0[8]} + {r0[9]}) / {n} = {mean1}}\\]
\nThe variance is given by the formula:
\n\\[\\textrm{Variance} = \\frac{1}{n}\\left(\\sum_{i=1}^{n}(x_i -\\mu)^2\\right)\\]
\nwhere the $x_i$ are the exam scores for {exam1}, $n=\\var{n}$ the number of students and $\\mu=\\var{mean1}$ the sample mean.
\nThe Standard Deviation is then the square root of the Variance i.e.
\nStandard Deviation = $\\sqrt{\\var{var1}} = \\var{stdev1}$ to one decimal place.
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