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practice 6 for terver added
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#import "@preview/elembic:1.1.0" as e
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#import "@preview/tiptoe:0.3.1"
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#import "@preview/lilaq:0.5.0" as lq
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// see: https://lilaq.org/docs/examples/schoolbook-style
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#show: lq.set-tick(inset: 1.5pt, outset: 1.5pt, pad: 0.4em)
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#show: lq.set-spine(tip: tiptoe.stealth)
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#show: e.show_(
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lq.label.with(kind: "y"),
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it => place(bottom + right, dy: -100% - .2em, dx: 50%+3.9em, rotate(90deg, it))
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)
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#show: e.show_(
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lq.label.with(kind: "x"),
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it => place(left + top, dx: 100% - .9em, dy: -3em, it)
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)
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#let x = lq.linspace(10, 25, num: 6)
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#let y = lq.linspace(110, 250, num: 8)
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//#let xc = lq.linspace(-0.1,0.1, num:100)
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#lq.diagram(
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xlabel: $x$,
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ylabel: $y$,
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width: 300pt,
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height: 300pt,
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xlim: (0, 25 + 5),
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ylim: (0, 250 + 20),
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xaxis: (position: 0),
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yaxis: (position: 0),
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lq.scatter(
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(10,10,10) + (13,13,13) + (16,16,16) + (19,19,19) + (22,22,22) + (25,25,25),
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(110,130,150) + (130,150,170) + (150,170,190) + (150,170,190) + (190,210,230) + (210,230,250),
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size: 5pt,
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color: red
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),
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lq.line(
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(0, 67.18), (100, 691.18)
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)
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)
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#let frame(stroke) = (x, y) => (
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left: if (x > 2 and x < 10) { 0pt } else { stroke },
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right: stroke,
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top: if (y < 3 or y > 7) { stroke } else { 0pt },
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bottom: stroke,
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)
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#align(center)[
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#rotate(-90deg, reflow: true)[
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#table(
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columns: 15,
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rows: 13,
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inset:(x: 4pt, y: 10pt),
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align: horizon,
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stroke: frame(1pt + rgb("21222C")),
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"", $j$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $11$, $12$, $13$,
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$i$, "X\\Y", $110$, $130$, $150$, $170$, $190$, $210$, $230$, $250$, $m_(x i)$, $m_(x i) x_i$, $sum^k_(j=1) m_(y j) y_j$, $x^2_i m_(x i)$, $x_i sum^k_(j=1) m_(i j) y_j$,
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$1$, $10$, $1$, $3$, $4$, $-$, $-$, $-$, $-$, $-$, $8$, $80$, $1100$, $800$, $11000$,
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$2$, $13$, $-$, $5$, $6$, $5$, $-$, $-$, $-$, $-$, $16$, $208$, $2400$, $2704$, $31200$,
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$3$, $16$, $-$, $-$, $4$, $8$, $6$, $-$, $-$, $-$, $18$, $288$, $3100$, $4608$, $49600$,
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$4$, $19$, $-$, $-$, $6$, $15$, $9$, $-$, $-$, $-$, $30$, $570$, $5160$, $10830$, $98040$,
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$5$, $22$, $-$, $-$, $-$, $-$, $5$, $6$, $7$, $-$, $18$, $396$, $3820$, $8712$, $84040$,
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$6$, $25$, $-$, $-$, $-$, $-$, $-$, $1$, $7$, $2$, $10$, $250$, $2320$, $6250$, $58000$,
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$7$, $m_x$, $1$, $8$, $20$, $28$, $20$, $7$, $14$, $2$, $100$, $1792$, $17900$, $33904$, $331880$,
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$8$, $m_(y j) y_j$, $110$, $1040$, $3000$, $4760$, $3800$, $1470$, $3320$, $500$, $17900$, $-$, $-$, $-$, $-$,
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$9$, $sum_(i=1)^6 m_(i j) x_i$, $10$, $95$, $296$, $478$, $377$, $157$, $329$, $50$, $1792$, $-$, $-$, $-$, $-$,
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$10$, $y^2_j m_(i j)$, $12100$, $135200$, $450000$, $809200$, $722000$, $308700$, $740600$, $125000$, $3 302 800$, $-$, $-$, $-$, $-$,
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$11$, $y_j sum_(i=1)^m m_(i j) x_i$, $1100$, $12350$, $44400$, $81260$, $71630$, $32970$, $75670$, $12500$, $331 880$, $-$, $-$, $-$, $-$,
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)
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]
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]
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#align(center)[
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#text(size:12pt)[
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Федеральное государственное автономное образовательное учреждение высшего
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образования «Национальный исследовательский университет ИТМО»\
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Дисциплина «Теория вероятностей»\
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]
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]
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#align(center + horizon)[
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#box(
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width: 100%,
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height: 8cm
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)[
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#text(size: 14pt)[
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*Практическая работа №6*\
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Вариант №12
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]
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]
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]
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#align(right)[
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Выполнил:\
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Решетников С.Е.\
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Проверила:\
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Селина Е.Г.
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]
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#align(center + bottom)[
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Дата сдачи: 19 ноября\
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Санкт-Петербург, 2025
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]

teorver/sem3/practice6/report.pdf

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teorver/sem3/practice6/report.typ

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#set page(
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margin: 1.5cm,
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numbering: "1"
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)
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#set par(justify: true)
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#set text(
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lang: "ru",
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size: 12pt,
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font: "New Computer Modern",
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)
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#show math.cases: math.display
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#show link: it => underline(text(fill: dark_blue)[#it])
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#include "pages/titlePage.typ"
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#pagebreak()
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Для подсчета числовых характеристик (выборочных средних $overline(x)$ и $overline(y)$, выборочных средних квадратичных отклонений $s_x$
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и $s_y$ и выборочного корреляционного момента $s_(x y)$) составляем рассчётную таблицу. При заполнении таблицы осуществляем контроль по строкам и столбцам:
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$
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sum_(i=1)^6 m_(x i) = sum_(j=1)^8 m_(y j) = n = 100\
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sum_(i=1)^6 sum_(j=1)^8 m_(i j) x_i = sum_(i=1)^6 m_(x_i) x_i = 1792\
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sum_(i=1)^6 sum_(j=1)^8 m_(i j) y_j = sum_(j=1)^8 m_(y j) y_j = 17900\
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sum_(i=1)^6(x_i sum_(j=1)^8 m_(i j) y_j) = sum_(j=1)^8 (y_j sum_(i=1)^6 m_(i j) x_i) = 331880
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$
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Вычисляем выборочные средние $overline(x)$ и $overline(y)$, $i=overline("1,6")$; $j=overline("1,8")$
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$
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overline(x) = (sum sum m_(i j) x_i)/n = (sum m_(x i) x_i)/n = 1792/100 = 17.92 ;\
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overline(y) = (sum m_(y i) y_i)/n = 17900/100 = 179.
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$
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Выборочные дисперсии находим по формулам:
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$
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s_x^2 = 1/(n-1) (sum m_(x i) x_i^2 - 1/n (sum m_(x i) x_i)^2 ) = 1/99 (33904 - 1/100 (1792)^2) approx 18.09\
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s_y^2 = 1/(n-1) (sum m_(y j) y_j^2 - 1/n (sum m_(y j) y_j)^2 ) = 1/99 (3302800 - 1/100 (17900)^2) approx 996.97
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$
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Корреляционный момент вычисляем по формуле:
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$
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s_(x y) = 1/(n-1)(sum sum m_(i j) x_i y_j - 1/n (sum m_(x i) x_i)(sum m_(y j) y_j))=\ = 1/99 (331880 - 1/100 (1792 dot 17900)) approx 112.24
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$
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Оценкой теоретической линии регрессии является эмпирическая линия регрессии, уравнение которой имеет вид
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$
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y = overline(y) + r_(x y) s_y/s_x (x - overline(x)) space "где"\
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s_x = sqrt(18.09) approx 4.25 quad quad quad
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s_y = sqrt(996.97) approx 31.57\
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r_(x y) = s_(x y) / (s_x s_y) = 112.24/(4.25 dot 31.57) approx 0.84
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$
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Составляем уравнение эмпирической линии регрессии $y$ на $x$
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$
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y = 179 + 0.84 dot 31.57/4.25 (x - 17.92) approx 6.24 dot x + 179 - 111.82 = 6.24 dot x + 67.18
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$
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#pagebreak()
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#include "pages/table.typ"
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#pagebreak()
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Строим линию регрессии и случайные точки $(x_i; y_j)$
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#align(center)[
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#include "pages/plot.typ"
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]

teorver/sem3/practice6/task.png

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