C02 - Random Variables
coin3 <- data.frame(
N_Heads = c(0, 1, 2, 3),
C_Heads = c(1, 3, 3, 1)
)
attach(coin3)
coin3$P_Heads/sum(C_Heads)
#options(repr.plot.width=7, repr.plot.height=4)
#barplot(coin3$P_Heads, names = coin3$N_Heads, main = "Discrete Probability Distribution of flipping three coins")
When attempting to create a histogram of the above data, an error stating "'height' must be expressed as a vector or matrix" occurs. I am unsure if this step is required for the assignment, but the command used in the example does not seem to function.
normal_dist <- function(x, b){
mu = mean(x)
sigma = sd(x)
x1 <- seq(mu - 6*sigma, mu + 6*sigma, length = 100)
# Normal curve
fun <- dnorm(x1, mean = mu, sd = sigma)
# Histogram
options(repr.plot.width=7, repr.plot.height=4)
hist(x, prob = TRUE, col = "white", breaks = b,
xlim = c(mu - 6*sigma, mu + 6*sigma),
ylim = c(0, max(fun)),
main = "Histogram overlayed with normal curve")
lines(x1, fun, col = 2, lwd = 2)
}
data(chickwts)
attach(chickwts)
normal_dist(chickwts$weight, 10)
shapiro.test(chickwts$weight)
The data can be approximated as normal distribution since p > 0.21.
chickwts$zscore = (weight-mean(weight))/sd(weight)
tail(chickwts)
Problem From The Book
GRE scores, Part I. Sophia who took the Graduate Record Examination (GRE) scored 160 on the Verbal Reasoning section and 157 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section for all test takers was 151 with a standard deviation of 7, and the mean score for the Quantitative Reasoning was 153 with a standard deviation of 7.67. Suppose that both distributions are nearly normal.
N_verbal = (μ=151,σ=7), N_quantitativereasoning = (μ=153,σ=7.67)
z_sophiaV = (160 - 151)/7
= 1.29, z_sophiaQ = (157 - 153)/7.67 = 0.52
Sophia scored 1.29 standard deviations above the mean on the Verbal Reasoning section and 0.52
standard deviations above the mean on the Quantitative Reasoning section.
Sophia performed better on the Verbal Reasoning section, as her z score was higher.
Sofia's verbal ability percentile score is 90.15 %, meaning 9.85 % of participents performed better than Sophia in verbal ability.
Sofia's quantitative reasoning percentile score is 69.85 %
, meaning 30.15 % of participants perfomed better than Sophia in quantitative reasoning.
We cannot compare the raw scores since they are on different scales. Sofia's scores will be
measured relative to the other students, hence Z scores can be helpful. Comparing percentile scores is a more appropriate way of determining how well
she performed compared to other participants.
Z scores can be calculated for distributions that are
not normal, so answer (b) would not change. However, we could not answer questions (c) through (f) since we cannot use the Z table to
calculate probabilities and percentiles without a normal model.