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Zadaniem studenta jest wyznaczenie miar pozycyjnych dla losowego zbioru 32 warto\u015bci.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Na podstawie danych w tabeli odpowiedz na podane poni\u017cej pytania. W odpowiedziach podaj warto\u015bci bez zaokr\u0105glania.
\n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n$\\var{r1[8]}$ | \n$\\var{r1[9]}$ | \n$\\var{r1[10]}$ | \n$\\var{r1[11]}$ | \n$\\var{r1[12]}$ | \n$\\var{r1[13]}$ | \n$\\var{r1[14]}$ | \n$\\var{r1[15]}$ | \n
$\\var{r1[16]}$ | \n$\\var{r1[17]}$ | \n$\\var{r1[18]}$ | \n$\\var{r1[19]}$ | \n$\\var{r1[20]}$ | \n$\\var{r1[21]}$ | \n$\\var{r1[22]}$ | \n$\\var{r1[23]}$ | \n$\\var{r1[24]}$ | \n$\\var{r1[25]}$ | \n$\\var{r1[26]}$ | \n$\\var{r1[27]}$ | \n$\\var{r1[28]}$ | \n$\\var{r1[29]}$ | \n$\\var{r1[30]}$ | \n$\\var{r1[31]}$ | \n
If you sort the data in increasing order you get the following table:
\n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n$\\var{r1[8]}$ | \n$\\var{r1[9]}$ | \n$\\var{r1[10]}$ | \n$\\var{r1[11]}$ | \n$\\var{r1[12]}$ | \n$\\var{r1[13]}$ | \n$\\var{r1[14]}$ | \n$\\var{r1[15]}$ | \n
$\\var{r1[16]}$ | \n$\\var{r1[17]}$ | \n$\\var{r1[18]}$ | \n$\\var{r1[19]}$ | \n$\\var{r1[20]}$ | \n$\\var{r1[21]}$ | \n$\\var{r1[22]}$ | \n$\\var{r1[23]}$ | \n$\\var{r1[24]}$ | \n$\\var{r1[25]}$ | \n$\\var{r1[26]}$ | \n$\\var{r1[27]}$ | \n$\\var{r1[28]}$ | \n$\\var{r1[29]}$ | \n$\\var{r1[30]}$ | \n$\\var{r1[31]}$ | \n
Denote the ordered data by $x_j$, thus $x_{10}=\\var{r1[9]}$ for example.
\nMinimum value: The minimum value is $x_1=\\var{r1[0]}$.
\nLower Quartile: As there is an even number of values, the Lower Quartile will lie between two values. Its position is calculated by finding
\n\\[\\frac{n+1}{4}=\\frac{\\var{n+1}}{4}=8\\frac{1}{4}\\]
\nHence the Lower Quartile lies between the 8th and 9th entries in the ordered table, so it is:
\n\\[0.75\\times x_8+0.25\\times x_9 = 0.75\\times\\var{r1[7]}+0.25\\times \\var{r1[8]}=\\var{lquartile}\\]
\nMedian: The position of the median in the table is given by
\n\\[ \\frac{2(n+1)}{4} = \\frac{\\var{2*(n+1)}}{4} = 16 \\frac{1}{2}\\]
\nThe median lies between the 16th and 17th entries in the ordered table and is given by:
\n\\[0.5\\times x_{16}+0.5\\times x_{17} = 0.5\\times\\var{r1[15]}+0.5\\times \\var{r1[16]}=\\var{median}\\]
\nUpper Quartile: As there is an even number of values, the Upper Quartile will lie between two values. Its position is calculated by finding
\n\\[\\frac{3(n+1)}{4}=\\frac{\\var{3*(n+1)}}{4}=24\\frac{3}{4}\\]
\nHence the Upper Quartile lies between the 24th and 25th entries in the ordered table.
\nWe find it is \\[0.25\\times x_{24}+0.75\\times x_{25} = 0.25\\times\\var{r1[23]}+0.75\\times \\var{r1[24]}=\\var{uquartile}\\]
\nMaximum value: The maximum value is $x_{32}=\\var{r1[31]}$
\nb)
\nThe range is defined to be
Range = Maximum – Minimum
and so in this case we have:
Range = $\\var{r1[31]}-\\var{r1[0]}=\\var{range}$.
The interquartile range is defined to be
\n\\[ \\text{Upper Quartile} – \\text{Lower Quartile} \\]
\nand so in this case we have:
\n\\[ \\text{Interquartile range} = \\var{uquartile}-\\var{lquartile}=\\var{uquartile-lquartile} \\]
\nMost of the data should be spanned by $4s$ where $s$ is the sample standard deviation.
\nThe range of values is $\\var{r1[31]}-\\var{r1[0]}=\\var{r1[31]-r1[0]}$ and so $s$ should be approximately
\n\\[ \\simplify[std]{({r1[31]}-{r1[0]}) / 4 = {(r1[31] -r1[0]) / 4}} \\]
\nThe most likely value for the sample standard deviation of the options presented is $\\var{guess1}$.
\n(The actual value is $\\var{stdev}$ to 2 decimal places).
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\nMinimum (xmin) | \nPierwszy kwartyl (Q1) | \nMediana (Me) | \nTrzeci kwartyl (Q3) | \nMaximum (xmax) | \n
---|---|---|---|---|
[[0]] | \n[[1]] | \n[[2]] | \n[[3]] | \n[[4]] | \n
Oblicz rozst\u0119p:
\nR= [[0]]
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\nRQ= [[0]]
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[[0]]${<=} {x_{typ}}<=$[[1]]