Question

Q6. Consider a random variable x following a continuous distribution. Suppose at x = 2.5, the left tail probability is 0.76. Compute the probability of all values of x that are more extreme than 2.5. Show details of your computation.

Q7. Consider the setting of Q6. Suppose we are also told that, at x = −0.8, the right tail probability is 0.86. Determine which one of 2.5 and −0.8 is a more extreme value for x. Show details of your answer.

Q8. Explain why our calculations for Q6 and Q7 will no longer hold if we remove the word "continuous" in the statement of Q6.

Please do questions 6-8

Answer 1

For question 6, the probability of values more extreme than x = 2.5 is 0.24. For question 7, the value x = -0.8 is more extreme as it has a higher right tail probability of 0.86 compared to the left tail probability's complement at x = 2.5. In question 8, calculations change because, unlike continuous distributions, discrete distributions have non-zero probabilities at exact values.

Step-by-step explanation:

Probability Calculations for Continuous Random Variables

For question 6, given the left tail probability at x = 2.5 is 0.76, the probability of all values more extreme than 2.5 (i.e., P(x > 2.5)) is the complement of this probability, which is 1 - 0.76 = 0.24.

For question 7, we are comparing the tail probabilities at x = 2.5 and x = -0.8. The more extreme value will have the higher tail probability. Given that the left tail probability at x = 2.5 is 0.76 and the right tail probability at x = -0.8 is 0.86, it appears that x = -0.8 is the more extreme value because 0.86 > 0.24 (the complement of the left tail probability at x = 2.5).

In question 8, the concept of tail probabilities only applies to continuous distributions, because for discrete distributions, the probability at individual points is non-zero. Thus, if we remove the word "continuous", our calculations assuming probabilities of exact values would be incorrect because for discrete random variables, P(x = c) ≠ 0.