The Geometric Distribution - AP Statistics

1. What type of variable is used in a geometric distribution?

   Discrete random variable. Counts whole number trials, not continuous values.

2. What is the skewness of a geometric distribution?

   Positively skewed. Most probability mass at $X=1$, decreasing for higher values.

3. How does the probability of success $p$ affect the mean of a geometric distribution?

   As $p$ increases, the mean decreases. Higher $p$ means success comes sooner on average.

4. Find $P(X \leq 4)$ for a geometric distribution with $p = 0.3$.

   $P(X \leq 4) = 1 - (0.7)^4$. Uses CDF formula with success probability $p=0.3$.

5. What is the cumulative distribution function of a geometric distribution?

   $P(X \leq k) = 1 - (1-p)^k$. CDF gives probability of success by trial $k$.

6. Find $P(X > 3)$ for a geometric distribution with $p = 0.2$.

   $P(X > 3) = (1-0.2)^3$. Probability of needing more than 3 trials equals $(0.8)^3$.

7. For $p = 0.25$, what is the mean $E(X)$ of the distribution?

   $E(X) = \frac{1}{0.25} = 4$. Expected value calculation using mean formula.

8. Which parameter defines a geometric distribution?

   Probability of success $p$. Single parameter $p$ completely specifies the distribution.

9. Calculate $P(X \leq 5)$ for a geometric distribution with $p = 0.2$.

   $P(X \leq 5) = 1 - (0.8)^5$. CDF calculation for success within 5 trials.

10. What happens to $Var(X)$ as $p$ increases in a geometric distribution?

    Variance decreases. Higher success probability reduces variability in waiting time.

11. What condition must be met for a geometric distribution to be applicable?

    Trials are independent with constant probability $p$ of success. Each trial must have same success chance and be unaffected by others.

12. Find the probability of the first success on or before the 3rd trial with $p = 0.5$.

    $P(X \leq 3) = 1 - (0.5)^3$. CDF formula gives probability of success by trial 3.

13. What is the standard deviation of a geometric distribution with $p = 0.4$?

    $\frac{\sqrt{1-0.4}}{0.4}$. Standard deviation equals $\sqrt{\frac{1-p}{p^2}}$.

14. Identify the key property of a geometric distribution related to memory.

    Memoryless property. Previous failures don't influence future trial outcomes.

15. What is the formula for the probability mass function of a geometric distribution?

    $P(X = k) = (1-p)^{k-1}p$. Standard PMF where $k$ is trial number and $p$ is success probability.

16. Find $P(X = 3)$ if $p = 0.4$ in a geometric distribution.

    $P(X = 3) = (0.6)^2 \times 0.4$. Two failures $(0.6)^2$ followed by success $0.4$.

17. If $X \sim \text{Geom}(0.25)$, find $P(X = 2)$.

    $P(X = 2) = (1-0.25)^1 \times 0.25$. One failure $(0.75)^1$ then success $0.25$ on trial 2.

18. Is the geometric distribution memoryless? Yes or No.

    Yes. Past failures don't affect future success probabilities.

19. What is the mode of a geometric distribution?

    Mode = 1. First trial always has highest probability of success.

20. Find the probability of the first success on the 4th trial with $p = 0.3$.

    $P(X = 4) = (0.7)^3 \times 0.3$. Three failures $(1-0.3)^3$ then success $0.3$ on trial 4.

21. State the relationship between geometric and negative binomial distributions.

    Geometric is a special case of negative binomial with $r=1$. Negative binomial generalizes to $r$ successes; geometric has $r=1$.

22. If $p = 0.3$, find $P(X > 5)$ for a geometric distribution.

    $P(X > 5) = (0.7)^5$. Complement of CDF gives probability of exceeding 5 trials.

23. What does the geometric distribution assume about trial independence?

    Trials are independent. Each trial outcome doesn't affect subsequent trial probabilities.

24. If trial success probability is 0.15, what is $E(X)$?

    $E(X) = \frac{1}{0.15}$. Mean equals $\frac{1}{p}$ for geometric distribution.

25. Determine $Var(X)$ for a geometric distribution with $p = 0.4$.

    $Var(X) = \frac{1-0.4}{0.4^2}$. Uses standard variance formula $\frac{1-p}{p^2}$ with $p=0.4$.

26. If $X \sim \text{Geom}(0.25)$, find $P(X = 2)$.

    $P(X = 2) = (1-0.25)^1 \times 0.25$. One failure $ $(0.75)^1$ $ then success $ 0.25 $ on trial 2.

27. What type of trials are used in a geometric distribution?

    Bernoulli trials. Independent trials with constant success probability.

28. Identify the random variable $X$ in a geometric distribution.

    The number of trials until the first success. Counts trials needed to achieve one success in independent trials.

29. Calculate $E(X)$ for a geometric distribution with $p = 0.5$.

    $E(X) = \frac{1}{0.5} = 2$. Expected trials equals reciprocal of success probability.

30. Determine $Var(X)$ for $p = 0.2$ in a geometric distribution.

    $Var(X) = \frac{1-0.2}{0.2^2}$. Standard variance formula applied with given probability.

31. What does the geometric distribution model in terms of failures?

    Number of failures before the first success. Alternative interpretation: failures before achieving success.

32. What is the main application of the geometric distribution?

    Modeling the number of trials until the first success. Primary use case in probability and statistics applications.

33. Calculate the probability of the first success on the 1st trial with $p = 0.6$.

    $P(X = 1) = 0.6$. Success on first trial occurs with probability $p$.

34. Describe the shape of the probability mass function of a geometric distribution.

    Decreasing exponential shape. Highest probability at $X=1$, decreasing exponentially thereafter.

35. Express the cumulative probability $P(X > k)$ for a geometric distribution.

    $P(X > k) = (1-p)^k$. Survival function giving probability of exceeding $k$ trials.

36. What is the mean (expected value) formula for a geometric distribution?

    $E(X) = \frac{1}{p}$. Higher success probability means fewer expected trials needed.

37. State the formula for the variance of a geometric distribution.

    $Var(X) = \frac{1-p}{p^2}$. Standard variance formula derived from geometric distribution theory.

38. Which distribution models the number of trials until first success?

    The geometric distribution. Specifically models waiting time until first success occurs.

39. What is the probability of success in a geometric distribution?

    The constant probability $p$. Success probability remains fixed across all trials.

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