Ask Question
22 January, 19:26

variables. a. If P (X > c) = α, determine P (X ≤ c) in terms of α. b. Suppose that P (Y > c) = α/2 and P (Y c). Determine P (-c ≤ Y ≤ c) in terms of α. c. Suppose that P (-c ≤ T ≤ c) = 1 - α and also suppose that P (T c). Find P (T > c) in terms of α.

+2
Answers (1)
  1. 22 January, 19:38
    0
    a) P (X ≤ c) = 1 - a

    b) P (-c ≤ Y ≤ c) = 1 - 1.5a

    c) P (T > c) = a/2

    Explanation:

    Given:-

    - Let c > 0 and 0 < a < 1.

    - Let X, Y & T be random variables.

    Find:-

    If P (X > c) = α, determine P (X ≤ c) in terms of α.

    Solution:-

    - To find the probability about the same constant "c" but with inverted signs " >" to "≤ " we will make use of total probability i. e 1 and subtract the given probability of any sign to get the probability with respect to other sign.

    P (X ≤ c) = 1 - P (X > c)

    P (X ≤ c) = 1 - a

    Find:-

    b. Suppose that P (Y > c) = α/2 and P (Y c).

    Determine P (-c ≤ Y ≤ c) in terms of α.

    Solution:-

    - To find the probability between the same limit "c" we will break the relation into two parts P (Y < - c) and P (Y ≤ c)

    P (-c ≤ Y ≤ c) = P (Y ≤ c) - P (Y < - c)

    P (-c ≤ Y ≤ c) = (answer to part a) - P (Y > c)

    P (-c ≤ Y ≤ c) = (1 - a) - a/2

    P (-c ≤ Y ≤ c) = 1 - 1.5a

    Find:-

    c. Suppose that P (-c ≤ T ≤ c) = 1 - α and also

    suppose that P (T c). Find P (T > c) in terms of α.

    Solution:-

    - To find the probability between the same limit "c" we will break the relation into two parts P (T < - c) and P (T ≤ c)

    P (-c ≤ T≤ c) = P (T ≤ c) - P (T < - c)

    1 - a = P (T ≤ c) - P (T > c)

    1 - a = [1 - P (T > c) ] - P (T > c)

    - a = - 2 P (T > c)

    P (T > c) = a/2
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “variables. a. If P (X > c) = α, determine P (X ≤ c) in terms of α. b. Suppose that P (Y > c) = α/2 and P (Y c). Determine P (-c ≤ Y ≤ c) in ...” in 📙 Computers & Technology if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers