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https://github.com/szymon-jozef/ZUT-notatki.git
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202 lines
3.5 KiB
R
202 lines
3.5 KiB
R
# układ równań
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A = matrix(c(-9, 15, -14, -1, 13, 5, 18, 6, 8, 12, 10, -4, -1, 5, 8, 12), 4, 4)
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B = c(4, 298, 294, 120)
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solve(A, B)
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solve(A)
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solve(A)[1, 4]
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# wektory
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## Skonstruować wektor w trzecich potęg wielokrotności liczby 7 od 21 do 29903
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## następnie zliczyć, ile wśród elementów tego wektora jest liczb, których dwie ostatnie cyfry to
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## 23.
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w = seq(21, 29903, 7)^3
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summary.factor(w %% 100 == 23)
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## Skonstruować wektor w kwadratów liczb od 1 do 100 a następnie zliczyć, które
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## cyfry oraz jak często występują na pozycji jedności w wyznaczonych kwadratach (użyć operatora modulo, funkcji summary oraz typu factor). Ile wśród kwadratów liczb od 1 do 100 jest
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## podzielnych przez 2, przez 3 oraz przez 2 i 3?
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w = (1:100)^2
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summary.factor(w %% 2 == 0)
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summary.factor(w %% 3 == 0)
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summary.factor(w %% 3 == 0 & w %% 2 == 0)
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# dwumianowy i geom | lista 2
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# zad 1
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# P(X = 2)
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dbinom(2, 4, 1/3)
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# P(X < 2)
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pbinom(1, 4, 1/3)
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# P(X > 2)
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pbinom(2, 4, 1/3, F)
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# zad 2
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prob = 0.02
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dgeom(3, prob)
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pgeom(2, prob)
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# zad 3
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size = 100
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prob = 0.01
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# P(X = 2)
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dbinom(2, size, prob)
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dpois(2, size * prob)
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# P(2 < X < 6)
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sum(dbinom(2:6, size, prob))
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pbinom(3, size, prob, F)
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# zad 4
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dbinom(2, 4, 1/7)
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dbinom(3, 5, 1/2) > dbinom(2, 3, 1/2)
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# kwantyle
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qnorm(0.97)
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qchisq(0.975, 9)
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qt(0.95, 9)
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qf(0.995, 10, 23)
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# zad 2
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# d)
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# |x - 2| < 3
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# x - 2 < 3 || x - 2 > -3
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# x < 5 || x > -1
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pnorm(5, 1.5, 2) - pnorm(-1, 1.5, 2)
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# e)
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# |x| > 0.5
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# x > 0.5 || x < -0.5
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pnorm(-0.5, 1.5 ,2) + pnorm(0.5, 1.5, 2, F)
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# f)
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# -1 < 2x - 1 < 1
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# 0 < 2x < 2
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# 0 < x < 1
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pnorm(1, 1.5, 2) - pnorm(0, 1.5, 2)
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# g)
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# 3x - 1 > 2 || 3x -1 < -2
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# 3x > 3 || 3x < -1
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# x > 1 || 3 < -1/3
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pnorm(-1/3, 1.5, 2) + pnorm(1, 1.5, 2, F)
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# zad 3
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# 5x - 1 > 2.5 || 5x -1 < -2.5
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# 5x > 3.5 || 5x < -1.5
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pnorm(-1.5/5, -4, 3.5) + pnorm(3.5/5, -4, 3.5, F)
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# 4x - 2 <= 5.5 || 4x -2 >= -5.5
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# -5.5 < 4x - 2 < 5.5
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# -3.5 < 4x < 7.5
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pnorm(7.5/4, -4, 3.5) - pnorm(-3.5/4, -4, 3.5)
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# zad 4
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# P(X < 900)
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round(pnorm(900, 1000, 50) * 4000)
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qnorm(0.13, 1000, 50, F)
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# zad 5
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round(pnorm(7.1, 6.8, 0.3) * 30)
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qnorm(0.15, 6.8, 0.3)
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# zad 6
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# 22:05 < X < 22:10
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# 22 * 5/60 < X < 22 * 10/60
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pnorm(22 * 1/6 * 60, 15, 13) - pnorm(22 * 5/60 * 60, 15, 13)
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pnorm(22, 15, 13, F)
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# standaryzacja rozkładu normalnego
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# zad 1
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# P(X > 5) = 70/200
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sigma = 1
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m = 5 - sigma * qnorm(70/200, lower.tail = FALSE)
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round(pnorm(6, m, sigma) - pnorm(3, m, sigma), 4)
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# P(X < 6) = 0.8
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sigma = sqrt(4)
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m = 6 - sigma * qnorm(0.8)
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round(pnorm(6, m, sigma) - pnorm(3, m, sigma) ,4)
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# zad 2
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m = 5.2
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U = qnorm(0.121)
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sigma = (3 - m)/U
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qnorm(0.8, m, sigma)
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# zad 3
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m = -5
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X = -2
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U = qnorm(0.95)
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sigma = (X - m)/U
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pnorm(0, m, sigma, F)
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# zad 4
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m = 0.46
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X = 0.5
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U = qnorm(0.6)
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sigma = (X - m)/U
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pnorm(0.7, m, sigma, F)
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# zad 1
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A = matrix(c(1, 1, qnorm(3/10), qnorm(3/4)), 2, 2)
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B = c(5281.21, 9090.9)
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wynik = solve(A, B)
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m = wynik[1]
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sigma = wynik[2]
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pnorm(7800, m, sigma, F) * 100
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qnorm(0.15, m, sigma)
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# zad 3
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m = 165.5
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# 155 < X < 176
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U = qnorm((1 - 0.93)/2)
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sigma = (155 - m)/U
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round(pnorm(170, m, sigma, F) * 100)
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# zad 4
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A = matrix(c(1, 1, qnorm(1/12, lower.tail = FALSE), qnorm(2/12)), 2, 2)
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B = c(0.9, 0.5)
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wynik = solve(A, B)
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m = wynik[1]
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sigma = wynik[2]
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qnorm(0.1, m, sigma, F)
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A = matrix(c(1, 1, qnorm(11/12), qnorm(2/12)), 2, 2)
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B = c(0.9, 0.5)
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wynik = solve(A, B)
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m = wynik[1]
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sigma = wynik[2]
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qnorm(0.1, m, sig, F)
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# zadanie 2 lista 6
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n = 2000
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p = 0.002
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q = 1 - p
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sum(dbinom(6:12, n, p))
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if (n >= 50 & p < 10) {
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print("Rozkład poissona")
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sum(dpois(6:12, n * p))
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} else {
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print("Nie można użyć rozkładu poissona")
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}
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