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04 Random Variables and Probability Distributions Free Online FlashCards

Study 04 Random Variables and Probability Distributions with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

What is a random variable?

Back

A random variable assigns a numerical value to each outcome of a random process; uppercase letters such as XX denote the variable.

02
Front

How do discrete and continuous random variables differ?

Back

A discrete random variable has finitely or countably infinitely many possible values; a continuous random variable can take any value within an interval.

03
Front

What conditions must a discrete PMF satisfy?

Back

A valid PMF satisfies 0≤P(X=x)≤10\leq P(X=x)\leq 1 for every xx, and ∑xP(X=x)=1\sum_xP(X=x)=1.

04
Front

For the message distribution, what is P(X≥2)P(X\geq2)?

Back

Add the probabilities for 2 and 3: P(X≥2)=0.30+0.15=0.45P(X\geq2)=0.30+0.15=0.45.

05
Front

When is a binomial distribution appropriate?

Back

A binomial distribution counts successes in a fixed number of independent two-outcome trials with a constant success probability.

06
Front

How does a continuous PDF represent probability?

Back

For a continuous variable, probability is area under the PDF over an interval; the PDF value at one point is not itself a probability.

07
Front

What is P(X=c)P(X=c) for a continuous random variable?

Back

For any continuous random variable, the probability of one exact value is P(X=c)=0P(X=c)=0.

08
Front

What does a CDF measure?

Back

The CDF is F(x)=P(X≤x)F(x)=P(X\leq x), and interval probability is found by P(a<X<b)=F(b)−F(a)P(a<X<b)=F(b)-F(a).

09
Front

For X∼Uniform⁡(0,10)X\sim\operatorname{Uniform}(0,10), find P(2<X<6)P(2<X<6).

Back

The interval has width 6−2=46-2=4 and density 110\frac{1}{10}, so P(2<X<6)=4(110)=0.40P(2<X<6)=4\left(\frac{1}{10}\right)=0.40.

10
Front

What is the expected value of a discrete random variable?

Back

For a discrete variable, E(X)=∑xxP(X=x)E(X)=\sum_x xP(X=x). It is the long-run average value.

11
Front

What is the expected message count per hour?

Back

Using E(X)=∑xxP(X=x)E(X)=\sum_xxP(X=x), the message distribution has E(X)=0(0.20)+1(0.35)+2(0.30)+3(0.15)=1.40E(X)=0(0.20)+1(0.35)+2(0.30)+3(0.15)=1.40.

12
Front

What does variance measure?

Back

Variance is the probability-weighted average of squared deviations from the mean: Var⁡(X)=∑x(x−μ)2P(X=x)\operatorname{Var}(X)=\sum_x(x-\mu)^2P(X=x).