Probability & Statistics Ch.2 Section 4 Question 6

· Mohammad-Ali Bandzar

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let XExponential(3)X \sim \text{Exponential}(3). Compute each of the following:

  • a) P(0<X<1)P(0 < X < 1)
  • b) P(0<X<3)P(0 < X < 3)
  • c) P(0<X<5)P(0 < X < 5)
  • d) P(2<X<5)P(2 < X < 5)
  • e) P(2<X<10)P(2 < X < 10)
  • f) P(X>2)P(X > 2)

Solution

From Example 2.4.5 in the textbook:

P(aXb)=abλeλxdx=eλaeλbP(a \leq X \leq b) = \int_a^b \lambda e^{-\lambda x} \, dx = e^{-\lambda a} - e^{-\lambda b}

In this case λ=3\lambda = 3.

(a) P(0<X<1)P(0 < X < 1)

P(0<X<1)=e3×0e3×1=1e30.9502P(0 < X < 1) = e^{-3 \times 0} - e^{-3 \times 1} = 1 - e^{-3} \approx 0.9502

(b) P(0<X<3)P(0 < X < 3)

P(0<X<3)=e0e9=1e90.9999P(0 < X < 3) = e^{0} - e^{-9} = 1 - e^{-9} \approx 0.9999

(c) P(0<X<5)P(0 < X < 5)

P(0<X<5)=e0e15=1e151P(0 < X < 5) = e^{0} - e^{-15} = 1 - e^{-15} \approx 1

(d) P(2<X<5)P(2 < X < 5)

P(2<X<5)=e6e150.00248P(2 < X < 5) = e^{-6} - e^{-15} \approx 0.00248

(e) P(2<X<10)P(2 < X < 10)

P(2<X<10)=e6e300.00248P(2 < X < 10) = e^{-6} - e^{-30} \approx 0.00248

(f) P(X>2)P(X > 2)

From Example 2.4.5:

P(Xx)=eλxP(X \geq x) = e^{-\lambda x}

P(X>2)=e3×2=e60.00248P(X > 2) = e^{-3 \times 2} = e^{-6} \approx 0.00248