A random variable X is exponentially distributed with an expected value of 50.

Question

a-1. What is the rate parameter λ? (Round your answer to 3 decimal places.)



a-2. What is the standard deviation of X?



b. Compute P(44 ≤ X ≤ 56). (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)



c. Compute P(36 ≤ X ≤ 64). (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

Answer 1

a-1. To find the rate parameter λ, we use the formula for the expected value of an exponentially distributed variable:

E(X) = 1/λ

Given that the expected value E(X) is 50, we can solve for λ:

50 = 1/λ

λ = 1/50 = 0.02 (rounded to 3 decimal places)

Therefore, the rate parameter λ is 0.02.

a-2. The standard deviation of an exponentially distributed variable with rate parameter λ is given by the formula:

σ = 1/λ

Using the rate parameter λ = 0.02 we found in part a-1:

σ = 1/0.02 = 50

Therefore, the standard deviation of X is 50.

b. To compute P(44 ≤ X ≤ 56), we use the cumulative distribution function (CDF) of the exponential distribution. The CDF is given by:

CDF(x) = 1 - e^(-λx)

P(44 ≤ X ≤ 56) = CDF(56) - CDF(44)

= (1 - e^(-0.02*56)) - (1 - e^(-0.02*44))

= (1 - e^(-1.12)) - (1 - e^(-0.88))

≈ 0.4970

Therefore, P(44 ≤ X ≤ 56) is approximately 0.4970.

c. Similarly, to compute P(36 ≤ X ≤ 64), we use the CDF:

P(36 ≤ X ≤ 64) = CDF(64) - CDF(36)

= (1 - e^(-0.02*64)) - (1 - e^(-0.02*36))

= (1 - e^(-1.28)) - (1 - e^(-0.72))

≈ 0.5042

Therefore, P(36 ≤ X ≤ 64) is approximately 0.5042.