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Zadaniem studenta jest wyznaczenie miar pozycyjnych dla losowego zbioru 32 warto\u015bci.

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Na podstawie danych w tabeli odpowiedz na podane poni\u017cej pytania. W odpowiedziach podaj warto\u015bci bez zaokr\u0105glania.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{r1[0]}$$\\var{r1[1]}$$\\var{r1[2]}$$\\var{r1[3]}$$\\var{r1[4]}$$\\var{r1[5]}$$\\var{r1[6]}$$\\var{r1[7]}$$\\var{r1[8]}$$\\var{r1[9]}$$\\var{r1[10]}$$\\var{r1[11]}$$\\var{r1[12]}$$\\var{r1[13]}$$\\var{r1[14]}$$\\var{r1[15]}$
$\\var{r1[16]}$$\\var{r1[17]}$$\\var{r1[18]}$$\\var{r1[19]}$$\\var{r1[20]}$$\\var{r1[21]}$$\\var{r1[22]}$$\\var{r1[23]}$$\\var{r1[24]}$$\\var{r1[25]}$$\\var{r1[26]}$$\\var{r1[27]}$$\\var{r1[28]}$$\\var{r1[29]}$$\\var{r1[30]}$$\\var{r1[31]}$
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a)

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If you sort the data in increasing order you get the following table:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{r1[0]}$$\\var{r1[1]}$$\\var{r1[2]}$$\\var{r1[3]}$$\\var{r1[4]}$$\\var{r1[5]}$$\\var{r1[6]}$$\\var{r1[7]}$$\\var{r1[8]}$$\\var{r1[9]}$$\\var{r1[10]}$$\\var{r1[11]}$$\\var{r1[12]}$$\\var{r1[13]}$$\\var{r1[14]}$$\\var{r1[15]}$
$\\var{r1[16]}$$\\var{r1[17]}$$\\var{r1[18]}$$\\var{r1[19]}$$\\var{r1[20]}$$\\var{r1[21]}$$\\var{r1[22]}$$\\var{r1[23]}$$\\var{r1[24]}$$\\var{r1[25]}$$\\var{r1[26]}$$\\var{r1[27]}$$\\var{r1[28]}$$\\var{r1[29]}$$\\var{r1[30]}$$\\var{r1[31]}$
\n

Denote the ordered data by $x_j$, thus $x_{10}=\\var{r1[9]}$ for example.

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Minimum value: The minimum value is $x_1=\\var{r1[0]}$.

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Lower Quartile: As there is an even number of values, the Lower Quartile will lie between two values. Its position is calculated by finding

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\\[\\frac{n+1}{4}=\\frac{\\var{n+1}}{4}=8\\frac{1}{4}\\]

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Hence the Lower Quartile lies between the 8th and 9th entries in the ordered table, so it is:

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\\[0.75\\times x_8+0.25\\times x_9 = 0.75\\times\\var{r1[7]}+0.25\\times \\var{r1[8]}=\\var{lquartile}\\]

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Median: The position of the median in the table is given by

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\\[ \\frac{2(n+1)}{4} = \\frac{\\var{2*(n+1)}}{4} = 16 \\frac{1}{2}\\]

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The median lies between the 16th and 17th entries in the ordered table and is given by:

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\\[0.5\\times x_{16}+0.5\\times x_{17} = 0.5\\times\\var{r1[15]}+0.5\\times \\var{r1[16]}=\\var{median}\\]

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Upper Quartile: As there is an even number of values, the Upper Quartile will lie between two values. Its position is calculated by finding

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\\[\\frac{3(n+1)}{4}=\\frac{\\var{3*(n+1)}}{4}=24\\frac{3}{4}\\]

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Hence the Upper Quartile lies between the 24th and 25th entries in the ordered table.

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We 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}\\]

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Maximum value: The maximum value is $x_{32}=\\var{r1[31]}$

\n

b)

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The range is defined to be

Range = Maximum – Minimum

and so in this case we have:

Range = $\\var{r1[31]}-\\var{r1[0]}=\\var{range}$.

\n

c)

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The interquartile range is defined to be

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\\[ \\text{Upper Quartile} – \\text{Lower Quartile} \\]

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and so in this case we have:

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\\[ \\text{Interquartile range} = \\var{uquartile}-\\var{lquartile}=\\var{uquartile-lquartile} \\]

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d)

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Most of the data should be spanned by $4s$ where $s$ is the sample standard deviation.

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The range of values is $\\var{r1[31]}-\\var{r1[0]}=\\var{r1[31]-r1[0]}$ and so $s$ should be approximately

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\\[ \\simplify[std]{({r1[31]}-{r1[0]}) / 4 = {(r1[31] -r1[0]) / 4}} \\]

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The 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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Oblicz warto\u015bci miar wskazanych poni\u017cej:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Minimum (xmin)Pierwszy kwartyl (Q1)Mediana (Me)Trzeci kwartyl (Q3)Maximum (xmax)
[[0]][[1]][[2]][[3]][[4]]
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Oblicz rozst\u0119p:

\n

R= [[0]]

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Oblicz rozst\u0119p mi\u0119dzykwartylowy:

\n

RQ= [[0]]

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Wyznacz typowy obszar zmienno\u015bci wzgl\u0119dem miar pozycyjnych ($Me{+-}Q$):
[[0]]${<=} {x_{typ}}<=$[[1]]

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