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The following table shows {what}, $X$, {units} {forwhat}.
\nCalculate the relative percentage frequencies (to one decimal place for all).
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---|---|---|
$\\var{a[0]}\\le X \\lt \\var{a[1]}$ | \n$\\var{norm1[0]}$ | \n[[0]] | \n
$\\var{a[1]}\\le X \\lt \\var{a[2]}$ | \n$\\var{norm1[1]}$ | \n[[1]] | \n
$\\var{a[2]}\\le X \\lt \\var{a[3]}$ | \n$\\var{norm1[2]}$ | \n[[2]] | \n
$\\var{a[3]}\\le X \\lt \\var{a[4]}$ | \n$\\var{norm1[3]}$ | \n[[3]] | \n
$\\var{a[4]}\\le X \\lt \\var{a[5]}$ | \n$\\var{norm1[4]}$ | \n[[4]] | \n
$\\var{a[5]}\\le X \\lt \\var{a[6]}$ | \n$\\var{norm1[5]}$ | \n[[5]] | \n
$\\var{a[6]}\\le X \\lt \\var{a[7]}$ | \n$\\var{norm1[6]}$ | \n[[6]] | \n
Given a table of the number of days in which sales were between £x1000 and £(x+1)1000 find the relative percentage frequencies of these volume of sales.
\nrebelmaths
", "licence": "Creative Commons Attribution 4.0 International"}, "functions": {"revsort": {"type": "list", "language": "jme", "definition": "list(-1*vector(sort(list(-1*vector(a)))))", "parameters": [["a", "list"]]}}, "ungrouped_variables": ["a", "what", "freqdays", "daysopen", "things", "m", "forwhat", "units", "s", "num", "rel", "n1", "y", "freqdays1", "freqdays2", "r", "norm1"], "advice": "By summing the 'number of days' column we note that there were $\\var{daysopen}$ days in the year when sales took place.
\nSo to work out the relative percentage for a given sales interval, we must calculate $\\frac{\\text{number of days}}{\\var{daysopen}}\\times 100$
\n\n\nE.g. for the interval $\\var{a[0]}\\le X \\lt \\var{a[1]}$:
\nThere were $\\var{norm1[0]}$ days out of the $\\var{daysopen}$ when there were between $\\var{a[0]}$ and $\\var{a[1]}$ thousand pounds worth of sales (including $\\var{a[0]}$ thousand but not $\\var{a[1]}$ thousand) .
\nHence the relative frequency percentage for such sales is given by \\[100 \\times \\frac{\\var{norm1[0]}}{\\var{daysopen}}\\%=\\var{rel[0]}\\%\\] to one decimal place.
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