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Случайная величина $xi$ задана функцией распределения $F_xi (x) = cases(0 ", " x lt.eq 0, a sin 2 x ", " 0 lt x lt.eq pi/2, 1 ", " x gt pi/2)$ коэффициент $a$, плотность распределения. + +*Решение*: + +Найдем $a$: + +Чтобы была правостоянтной и непрерывной в точке $x = pi/2$: + +$ +a sin^2(pi/2) = a = 1 arrow.double a = 1. +$ + +Найдем плотность $f_xi (x)$ + +$ +f_xi (x) = F'(x) = cases(0 ", " x lt.eq 0, frac(d, d x) sin^2 x = 2 sin x cos x = sin(2 x) ", " 0 lt x lt pi/2, 0 ", " x lt.eq pi/2.) +$ + +*Ответ*: $a = 1, space f_xi (x) = cases(sin(2x) ", " 0 lt x lt pi/2, 0 ", иначе.")$ + +#line(length: 100%) + +*2*. Плотность распределения СВ $xi$ равна $f_xi (x) = cases(0 ", " x lt.eq 0 ", " x gt pi/4, a cos 2 x ", " 0 lt x lt.eq pi/4, 1 ", " x gt pi/2)$. Найти коэффициент $a$, функцию распределения и вероятность попадания в интервал $(0, pi/6)$. + +*Решение*: + +Коэффициент $a$: +$ +integral_0^(pi/4) a cos 2 x space d x = 1 arrow.double a dot frac(sin(pi/2), 2) = a/2 = 1 arrow.double a=2 +$ + +Функция распределения $F_xi (x)$: + +$ +F_xi (x) = cases(0 ", " x lt.eq 0, integral_0^x 2 cos 2 t space d t = sin 2 x ", " 0 lt x lt.eq pi/4, 1 ", " x gt pi/4) +$ + +Вероятность $P(0 lt xi lt pi/6)$: +$ +F_xi (pi/6) - F_xi (0) = sin(2 dot pi/6) = sin(pi/3) = frac(sqrt(3), 2). +$ + +*Ответ*: $a = 2, space F_xi (x) = cases(0 ", " x lt.eq 0, integral_0^x 2 cos 2 t space d t = sin 2 x ", " 0 lt x lt.eq pi/4, 1 ", " x gt pi/4), space P(0 lt xi lt pi/6) = frac(sqrt(3), 2). +$ + +#line(length: 100%) + +*3*. Плотность распределения СВ $xi$ равна $f_xi (x) = cases(0 " , " x lt.eq -1 " , " x gt 1, frac(a, sqrt(1 - x^2)) " , " -1 lt x lt.eq 1)$. Найти коэффициент $a$, функцию распределения и вероятность попадания в интервал $(0, 1/2)$. + +*Решение*: + +Коэффициент $a$: + +$ +integral_(-1)^1 frac(a, sqrt(1 - x^2)) d x = a[arcsin x]_(-1)^1 = a pi = 1 arrow.double a = 1/pi +$ + +Функция распределения: +$ +F_xi (x) = cases(0 ", " x lt.eq -1, frac(arcsin x + pi/2, pi) ", " -1 lt x lt.eq 1, 1 ", " x gt 1) +$ + +Вероятность $P(0 < xi < 1/2)$: +$ +F_xi (1/2) - F_xi(0) = 2/3 - 1/2 = 1/6 +$ + +*Ответ*: $a = 1/pi, space F_\xi(x) = cases(0 ", " x lt.eq -1, frac(arcsin x + pi/2, pi) ", " -1 lt x lt.eq 1, 1 ", " x gt 1), space P(0 lt xi lt 1/2) = 1/6$. + +#line(length: 100%) + +*4*. Автобусы некоторого маршрута идут строго по расписанию. Интервал движения 5 мин. Найти вероятность того, что пассажир, подошедший к остановке, будет ожидать автобус менее 3 мин. + +*Решение*: + +Пусть $xi$ — время ожидания автобуса. + +Так как пассажир приходит случайно, $xi$ равномерно распределена на интервале от 0 до 5 минут (от последнего автобуса до следующего): + +$ +xi tilde U(0, 5) +$ + +Плотность равномерного распределения на $[0,5]$: + +$ +f_xi (x) = 1/5, space.quad 0 lt.eq x lt.eq 5 +$ + +Функция распределения: + +$ +F_xi (x) = P(xi lt.eq x) = integral_0^x f_xi (t) space d t = x/5, space.quad 0 lt.eq x lt.eq 5 +$ + + +Находим вероятность + +Нужно $P(xi lt 3)$: + +$ +P(xi lt 3) = integral_0^3 f_xi (x) space d x = integral_0^3 1/5 space d x = 3/5 = 0.6 +$ + +*Ответ*: $0.6$. + +#line(length: 100%) + +*5*. Случайная величина $xi$ задана плотностью $f_xi (x) = 2 e^(-2x)$. Найти вероятность попадания в интервал $(1, 2)$. + +*Ответ*: $P(1 < xi < 2) = integral_1^2 2 e^(-2 x) space d x = [-e^(-2x)]_1^2 = e^(-2) - e^(-4) approx 0.117$. + +#line(length: 100%) + +*6*. Производится взвешивание некоторого вещества. Случайные ошибки взвешивания подчинены нормальному закону распределения с $a = 0, space sigma = 20 "г."$ Найдите вероятность того, что взвешивание будет произведено с ошибкой, не превосходящей, по модулю $10 "г."$ + +*Решение*: Для нормальной случайной величины вероятность через стандартное нормальное распределение $Z tilde N(0,1)$ вычисляется так: + +$ +Z = frac(xi - a, sigma) +$ + +Здесь $a = 0, sigma = 20$, значит: + +$ +Z = frac(xi, 20) +$ + +Перевод интервала + +Нам нужно: + +$ +P(-10 lt.eq xi lt.eq 10) = P ((-10)(20) lt.eq Z lt.eq (10)(20)) = P(-0.5 lt.eq Z lt.eq 0.5) +$ + +Используем функцию стандартного нормального распределения $Phi(z)$ + +$ +P(-0.5 lt.eq Z lt.eq 0.5) = Phi(0.5) - Phi(-0.5) +$ + +Свойство функции распределения: + +$ +Phi(-z) = 1 - Phi(z) +$ + +Тогда: + +$ +P(-0.5 lt.eq Z lt.eq 0.5) = Phi(0.5) - (1 - Phi(0.5)) = 2 Phi(0.5) - 1 +$ + +Находим численно + +Стандартные таблицы: + +$ +Phi(0.5) approx 0.6915 +$ + +Тогда: + +$ +P(-0.5 lt.eq Z lt.eq 0.5) = 2 dot 0.6915 - 1 = 1.383 - 1 = 0.383 +$ + +*Ответ*: $0.383$. + diff --git a/pracs/prac8/task.pdf b/pracs/prac8/task.pdf new file mode 100644 index 0000000..63ec0bb Binary files /dev/null and b/pracs/prac8/task.pdf differ