SD06 — Continuous Distributions
Course: Statistics and Probability for Data Science (DS14270).
Source deck: Samer El Kababji, SD06 — Continuous Distributions.
This lesson covers the normal density, standardization, probability calculations, the empirical rule, and linear combinations of normal variables.
If \(X\sim N(\mu,\sigma^2)\), then
\[ f(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right], \qquad -\infty<x<\infty. \]
Its mean is \(\mu\) and its variance is \(\sigma^2\).
mu <- 100
sigma <- 15
x <- seq(mu - 4 * sigma, mu + 4 * sigma, length.out = 501)
plot(
x, dnorm(x, mean = mu, sd = sigma), type = "l", lwd = 2,
xlab = "x", ylab = "f(x)", main = "N(100, 15^2) density"
)
abline(v = mu, lty = 2, col = "steelblue")
grid()
The standard normal variable \(Z\sim N(0,1)\) has density \(\phi(z)\) and CDF \(\Phi(z)=P(Z\leq z)\).
Standardization converts a normal variable to the standard normal scale:
\[ Z=\frac{X-\mu}{\sigma}. \]
A z-score tells how many standard deviations a value lies above or below the mean. With exam mean 70, standard deviation 10, and score 85,
\[ z=\frac{85-70}{10}=1.5. \]
score <- 85
mu <- 70
sigma <- 10
z <- (score - mu) / sigma
z[1] 1.5For any cutoff \(x\),
\[ P(X\leq x)=P\left(Z\leq\frac{x-\mu}{\sigma}\right). \]
In the deck, \(X\sim N(100,15^2)\). For \(x=120\),
\[ z=\frac{120-100}{15}\approx1.33, \]
so \(P(X\leq120)\approx0.9082\) when a z-table rounds to 1.33. R retains the unrounded z-score.
z_exact <- (120 - 100) / 15
c(
exact_z_in_R = pnorm(z_exact),
rounded_z_table = pnorm(1.33),
direct = pnorm(120, mean = 100, sd = 15)
) exact_z_in_R rounded_z_table direct
0.9087888 0.9082409 0.9087888 For a normal distribution, approximately 68%, 95%, and 99.7% of observations fall within 1, 2, and 3 standard deviations of the mean.
k <- 1:3
exact_probabilities <- pnorm(k) - pnorm(-k)
data.frame(
interval = paste0("mu +/- ", k, " sigma"),
exact_probability = exact_probabilities,
rule_of_thumb = c(0.68, 0.95, 0.997)
) interval exact_probability rule_of_thumb
1 mu +/- 1 sigma 0.6826895 0.680
2 mu +/- 2 sigma 0.9544997 0.950
3 mu +/- 3 sigma 0.9973002 0.997The \(p\)th quantile satisfies \(P(X\leq x_p)=p\) and is
\[ x_p=\mu+\sigma z_p, \qquad z_p=\Phi^{-1}(p). \]
p <- c(0.025, 0.50, 0.975)
data.frame(
probability = p,
standard_normal = qnorm(p),
N_100_15_squared = qnorm(p, mean = 100, sd = 15)
) probability standard_normal N_100_15_squared
1 0.025 -1.959964 70.60054
2 0.500 0.000000 100.00000
3 0.975 1.959964 129.39946If \(X\sim N(\mu,\sigma^2)\) and \(Y=aX+b\), then
\[ Y\sim N(a\mu+b,a^2\sigma^2). \]
If independent \(X_1\sim N(\mu_1,\sigma_1^2)\) and \(X_2\sim N(\mu_2,\sigma_2^2)\), then
\[ X_1+X_2\sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2). \]
mu_X <- 10
sd_X <- 3
a <- -2
b <- 5
c(
transformed_mean = a * mu_X + b,
transformed_sd = abs(a) * sd_X
)transformed_mean transformed_sd
-15 6 mu_1 <- 10; sd_1 <- 3
mu_2 <- 4; sd_2 <- 2
c(sum_mean = mu_1 + mu_2, sum_sd = sqrt(sd_1^2 + sd_2^2)) sum_mean sum_sd
14.000000 3.605551 pnorm(62, 50, 8) is approximately \(0.9332\); the central interval has probability pnorm(58, 50, 8) - pnorm(42, 50, 8), approximately \(0.6827\).qnorm(0.90, 50, 8) is approximately \(60.25\).