05 Common Statistical Distributions Free Online FlashCards

How do discrete and continuous distributions differ?

A discrete random variable takes countable values and uses a PMF; a continuous random variable takes values in an interval and uses a PDF. For a continuous variable, any single exact value has probability zero.

What assumptions define a binomial distribution?

The binomial model requires a fixed number of trials, two outcomes per trial, independent trials, and the same success probability on every trial.

What are the mean and variance of a binomial distribution?

If $X\sim\operatorname{Binomial}(n,p)$, then $E(X)=np$ and $\operatorname{Var}(X)=np(1-p)$. Its standard deviation is $\sqrt{np(1-p)}$.

What is the binomial PMF?

For $X\sim\operatorname{Binomial}(n,p)$, the probability of exactly $x$ successes is $P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$ for $x=0,1,\ldots,n$.

What does a geometric distribution model?

The geometric distribution models the number of independent Bernoulli trials required to obtain the first success, with constant success probability $p$.

What is the geometric PMF?

For $X\sim\operatorname{Geometric}(p)$, $P(X=x)=(1-p)^{x-1}p$. The first $x-1$ trials fail and the $x$-th trial succeeds.

What is the geometric distribution’s memoryless property?

The memoryless property is $P(X>s+t\mid X>s)=P(X>t)$. Previous failures do not change the probability of needing more than $t$ additional trials.

How is an interval probability found for $U(a,b)$?

For $X\sim U(a,b)$ and $[c,d]\subseteq[a,b]$, $P(c\le X\le d)=\frac{d-c}{b-a}$. Probability is proportional to interval length.

What are the mean and variance of $U(a,b)$?

For $X\sim U(a,b)$, $E(X)=\frac{a+b}{2}$ and $\operatorname{Var}(X)=\frac{(b-a)^2}{12}$.

What characterizes a normal distribution?

A normal distribution is continuous, symmetric, and bell-shaped. Its mean, median, and mode are equal, and it is determined by $\mu$ and $\sigma>0$.

What is the standard normal distribution?

The standard normal distribution has mean $0$ and standard deviation $1$: $Z\sim N(0,1)$. Its CDF is written $\Phi(z)=P(Z\le z)$.

How do you standardize a normal variable?

For $X\sim N(\mu,\sigma^2)$, standardize with $z=\frac{x-\mu}{\sigma}$. Then $P(X\le x)=\Phi\left(\frac{x-\mu}{\sigma}\right)$.