Final answer:
Calculations for different probabilities within a geometric distribution with a success rate of 0.3 are based on the formula for a geometric random variable. Examples include specific event probabilities and cumulative probabilities, as well as the distribution's mean and standard deviation.
Step-by-step explanation:
The student has asked about how to calculate different probabilities related to a geometric distribution with a probability of success of 0.3. The questions pertain to finding the probability of specific events occurring in this distribution.
- a. P(X = 6): To find this, we calculate (1 - 0.3)5 × 0.3, because the first five trials must be failures, followed by a success on the sixth trial.
- b. P(x ≤ 3): This is the sum of P(X = 1), P(X = 2), and P(X = 3).
- c. P(X > 4): To find this, we use 1 minus the probability of P(x ≤ 4).
- d. P(3 ≤ X ≤ 10): To calculate this, sum the probabilities for X = 3 through X = 10.
- e. P(3 < X < 6): This is the sum of the probabilities for X = 4 and X = 5, since we exclude the boundaries.
To answer the rest of the questions related to specific examples and to find the mean and standard deviation, the student should use the geometric distribution's mean and standard deviation formulas, with the mean equaling 1/p and the standard deviation as √((1 - p) / p2).