Aula 07
Pedro Gonçalves da Cunha
2023-05-17
Carregando bibliotecas e base de dados
library(RcmdrMisc)
library(readxl)
QE <- read_excel("C:\\Users\\15781634711\\Desktop\\Base_de_dados-master\\Questionario_Estresse.xls")Probabilidade
A distribuição normal
A curva normal, ou curva de sino, é a distribuição de probabilidade masi importante da estatística.
A área total do gráfico tem que ser igual a 1 para ser uma probabilidade.
Z ~ N(0, 1) (0 é a média e 1 é o desvio-padrão)
p(-1 ≤ Z ≤ 1) = 0,6826
p(-2 ≤ Z ≤ 2) = 0,9544
p(-3 ≤ Z ≤ 3) = 0,9974
p(-1 ≤ Z ≤ 2) = p1(-1 ≤ Z ≤ 0) + p2(0 ≤ Z ≤ 2) = 0,6826/2 + 0,9544/2 = 0,8185
p(1 ≤ Z ≤ 2) = p1(0 ≤ Z ≤ 2) - p2(0 ≤ Z 1) = 0,4772 - 0,3413 = 0,1359
p(-3 ≤ Z ≤ -2) = p1(-3 ≤ Z ≤ 0) - p2(-2 ≤ Z ≤ 0) = 0,0215
p(0 ≤ Z ≤ 1,34) = tabelado
Logo p(0 ≤ Z ≤ 1,34) = 0,40988
E p(-2,17 ≤ Z ≤ 1,11)?
x = seq(-3.291, 3.291, length.out=1000)
plotDistr(x, dnorm(x, mean = 0, sd = 1), cdf = FALSE, xlab = "x",
ylab = "Densidade", regions = list(c(-2.17, 0), c(0, 1.11)),
col = c('#0080C0', 'skyblue'), legend = FALSE)
Resposta: 0,8515
E p(1 ≤ Z ≤ 2,05)?
plotDistr(x, dnorm(x, mean = 0, sd = 1), cdf = FALSE, xlab = "x",
ylab = "Densidade", regions = list(c(0, 1), c(1, 2.05)),
col = c('red', 'darkred'), legend = FALSE)
Resposta: 0,13852
——————————————————————————————————–
Questões
1)
a)
Z~N(0,1)
p(-0,87 ≤ Z ≤ 1,54)
p = 0,30785 + 0,43822 = 0,74607
b) p(Z = 1,54) = 0
p(Z > 2,5) = 1 - pr(Z < 2,5) = 1 - (0,5 - pr(0 ≤ Z ≤ 2,5)) = 1 - (0,5 + 0,49379) = 0,00621
2)
a) A = N(50, 10)
pr(45 ≤ Z ≤ 65)
pr(45 - 50/10 ≤ Z ≤ 65-50/10)
pr(5/10 ≤ Z ≤ 15/10)
pr(0,5 ≤ Z ≤ 1,5)
pr = 0,19145 + 0,43319 = 0,62464
b)
pr(X > 57) = 1 - pr(Z < 57) = 0,5 - pr(0 ≤ Z ≤ 0,7)) = 0,5 - 0,25804
pr(X > 57) = 0,24196
3)
x = N(500, 100)
a) pr(480 ≤ z ≤ 530) = pr(-0,2 ≤ z ≤ 0,3) = 0,07926 + 0,11791 = 0,19717
b) pr(510 ≤ Z ≤ 540) = pr(0,1 ≤ Z ≤ 0,4) = 0,15542 - 0,03983 = 0,11559
c) pr(Z > 450) = 0,5 + pr(-0,5 ≤ Z ≤ 0) = 0,5 + 0,19146 = 0,69146
4)
w = N(10,10)
a) pr(w>3) = 0,5 + pr(0 ≤ w ≤ 0,7) = 0.5 + 0.25804 = 0.75804
b) pr(5 ≤ w ≤ 20) = pr(-0,5 ≤ w ≤ 1) = 0.19146 + 0.34134 = 0.5328
———————————————————————————————————
Fazendo as contas no R.
O R utiliza a tabela partindo do menos infinito.
pnorm(3, 10, 10, lower.tail = FALSE)## [1] 0.7580363pnorm(20, 10, 10) - pnorm(5, 10, 10)## [1] 0.5328072amostra_normal = rnorm(100)
amostra_normal## [1] -1.178769892 0.494370407 0.626334477 1.268233290 0.499311440
## [6] -0.716730655 -0.104566278 0.397216479 -2.035292273 -0.718963388
## [11] -0.261359689 0.713850080 1.561973850 -1.320435752 0.269599698
## [16] 1.180474294 -1.359804025 -0.377886072 1.813653121 -0.843158843
## [21] -0.656205213 0.489074652 0.575794957 0.216293003 1.699713165
## [26] -0.343288358 -0.311722211 0.536701899 -2.339614437 -0.563570620
## [31] -0.038864244 -2.015247853 -0.582493189 -0.013360164 -0.015558934
## [36] -2.042703809 -0.007760309 0.897852511 0.047315476 0.536439578
## [41] 1.486091234 3.562689304 0.726768828 -1.653629655 -0.978188275
## [46] -0.711072268 0.528993283 -0.927801381 -0.454371707 0.107189458
## [51] 0.843331502 -0.091986582 0.368739085 -0.926295926 -0.089159541
## [56] -0.278858491 0.446284218 -0.398352982 1.264085236 0.941278929
## [61] 0.049271850 0.075983375 -0.380081265 1.443308513 -0.913986956
## [66] 0.037786667 0.419084210 1.633497969 -0.026141014 -0.638535412
## [71] -1.961209944 -0.335818030 0.635071386 0.658385583 -0.775569288
## [76] -0.954738118 0.548291681 0.793407973 0.610292369 0.154278551
## [81] 1.226637327 2.849684935 0.001840172 0.771169509 -0.806336014
## [86] -1.807351035 -0.206077501 0.806584871 -1.027781910 -0.814646216
## [91] 0.438908384 0.644422915 -0.889243310 -0.751937514 0.298323672
## [96] 0.825495388 -1.207962266 -0.492464704 0.600338306 -0.274294043qqnorm(amostra_normal, col = "skyblue")
abline(a = 0, b = 1)
Testando a base de dados Questionario_estresse.
QE$Horas_padronizada = scale(QE$Horas_estudo)
qqnorm(QE$Horas_padronizada, col = "red")
abline(a = 0, b = 1)
QE$Estresse_padrao = scale(QE$Estresse)
qqnorm(QE$Estresse_padrao, col = "skyblue")
abline(a = 0, b = 1)
QE$Desempenho_padrao = scale(QE$Desempenho)
qqnorm(QE$Desempenho_padrao, col = "darkgreen")
abline(a = 0, b = 1)
hist(QE$Desempenho_padrao, col = c("darkgreen"))