Publications by aardvarkfunnyxia
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set.seed(12345) Generate 4 sets of 100 points from the N(0, 1) distribution random_normal_1 <- rnorm ( n = 100, mean = 0, sd = 1 ) random_normal_2 <- rnorm ( n = 100, mean = 0, sd = 1 ) random_normal_3 <- rnorm ( n = 100, mean = 0, sd = 1 ) random_normal_4 <- rnorm ( n = 100, mean = 0, sd = 1 ) compare each of the sets using histogram par (mfrow ...
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Exercise 8.1. x_values = seq ( -3, 3, length.out = 100000) plot ( x_values, dnorm (x_values, mean = 0, sd = 1), type = "l") abline ( v = c (-3, 3), col = "red", lty = 2, lwd = 2) total_area <- pnorm (Inf, mean = 0, sd = 1) area_below_3 <- pnorm (3, mean = 0, sd = 1) area_between <- ((2 * area_below_3 - 1) / total_area ) *100 print...
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Exercise 8.4 – an optional one set.seed(12345) Generate 4 sets of 100 points from the N(0, 1) distribution random_normal_1 <- rnorm ( n = 100, mean = 0, sd = 1 ) random_normal_2 <- rnorm ( n = 100, mean = 0, sd = 1 ) random_normal_3 <- rnorm ( n = 100, mean = 0, sd = 1 ) random_normal_4 <- rnorm ( n = 100, mean = 0, sd = 1 ) compare each of the...
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Using sample() in R allows you to take a random sample of elements from a dataset or a vector,either with or without replacement The basic syntax is the following sample (x, size, replace = FALSE, prob = NULL) (see ?sample) Generating a sample from a Vector Suppose we have a vector with 10 elements vector <- c (seq (1:10)) # to generate a random ...
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poi5 <- dpois(0:20, lambda = 5) plot (poi5, pch = 20, xlab = "value of x", ylab = "probability", main = "Lines of mean = 5", ylim = c(0,0.25)) lines(poi5, lwd = 2) ## the mean of the distribution Poi(5) mean_poi5 <- sum (0:20 * poi5) mean_poi5 ## [1] 4.999998 ## add the following lines (3): ### bin(10,0.5) distribution in red bin10_0....
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Exercise 7.1. ############################################### ### Expectation: (Note: in all of the following questions, p = 0.2, days = 365) days <- 365 p <- 0.2 E <- p * days print(E) ## [1] 73 ### Probability that I wear an ironed shirt 16 days or fewer: sum (dbinom (x = 1:16, size = days, prob = p)) ## [1] 4.096763e-18 Exercise 7.2. ###########...
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Exercise 7.1. ############################################### ### Expectation: (Note: in all of the following questions, p = 0.2, days = 365) days <- 365 p <- 0.2 E <- p * days print(E) ## [1] 73 ### Probability that I wear an ironed shirt 16 days or fewer: sum (dbinom (x = 1:16, size = days, prob = p)) ## [1] 4.096763e-18 Exercise 7.2. ###########...
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## Given that we have a normal distribution X ~ N (2, 4) show that P (X < 4) = 0.8413447 pnorm (4, mean= 2, sd = sqrt(4)) ## [1] 0.8413447 What are qX1 such that P(X < qX1) = 0.95 and qX2 such that P(X < qX2) = 0.975? qx1 <- qnorm (0.95, mean = 2, sd = sqrt(4)) qx1 ## [1] 5.289707 qx2 <- qnorm (0.975, mean = 2, sd = sqrt (4)) qx2 ## [1] 5.919928 Dr...
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Exercise 7.11. Create a QQ plot for the Handspan data with an appropriate heading. Do you think that the data appear normal? Why? qanda <- read.csv("selective_affinities.csv") qanda_modified <- read.csv("selective_affinities_aftermodified.csv") attach (qanda_modified) qqnorm (y = Handspan, pch = 20) qqline (Handspan, col = "red", lwd = 2) detach(...
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qanda <- read.csv("selective_affinities.csv") qanda_modified <- read.csv("selective_affinities_aftermodified.csv") attach(qanda_modified) par (mfrow = c (2,2)) # Histogram 1 - these comments break up the code hist (Handspan, freq = TRUE, main = "Handspan - breaks every 0.5cm", xlab = "handspan", ylab = "frequency", breaks = seq (floor...
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