Probability & Statistics Ch.2 Section 4 Question 3

· 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 Z∼Exponential(4)Z \sim \text{Exponential}(4). Compute each of the following:

  • a) P(Z≥5)P(Z \geq 5)
  • b) P(Z≥−5)P(Z \geq -5)
  • c) P(Z2≥9)P(Z^2 \geq 9)
  • d) P(Z4−17≥9)P(Z^4 - 17 \geq 9)

Solution

From Example 2.4.5 in the textbook:

P(X≥x)=∫x∞λe−λzdz=e−λxP(X \geq x) = \int_x^{\infty} \lambda e^{-\lambda z} dz = e^{-\lambda x}

In this case λ=4\lambda = 4.

(a) P(Z≥5)P(Z \geq 5)

P(Z≥5)=e−4×5=e−20≈2.06×10−9P(Z \geq 5) = e^{-4 \times 5} = e^{-20} \approx 2.06 \times 10^{-9}

(b) P(Z≥−5)P(Z \geq -5)

Since Example 2.4.5 says that f(x)=0f(x) = 0 for any x<0x < 0, we take the integral for domain [0,∞)[0, \infty), which is the same as P(Z≥0)P(Z \geq 0):

P(Z≥−5)=e−4×0=e0=1P(Z \geq -5) = e^{-4 \times 0} = e^0 = 1

(c) P(Z2≥9)P(Z^2 \geq 9)

If Z2≥9Z^2 \geq 9, then Z≥3Z \geq 3 or Z≤−3Z \leq -3.

For Z≥3Z \geq 3:

P(Z≥3)=e−12≈6.14×10−6P(Z \geq 3) = e^{-12} \approx 6.14 \times 10^{-6}

For Z≤−3Z \leq -3: Since the exponential distribution is only defined for x≥0x \geq 0:

P(Z≤−3)=0P(Z \leq -3) = 0

Total:

P(Z2≥9)=e−12+0=e−12≈6.14×10−6P(Z^2 \geq 9) = e^{-12} + 0 = e^{-12} \approx 6.14 \times 10^{-6}

(d) P(Z4−17≥9)P(Z^4 - 17 \geq 9)

Z4−17≥9Z^4 - 17 \geq 9

Z4≥26Z^4 \geq 26

Z≥264 or Z≤−264Z \geq \sqrt[4]{26} \text{ or } Z \leq -\sqrt[4]{26}

For Z≥264Z \geq \sqrt[4]{26}:

P(Z≥264)=e−4264≈0.000119P(Z \geq \sqrt[4]{26}) = e^{-4\sqrt[4]{26}} \approx 0.000119

For Z≤−264Z \leq -\sqrt[4]{26}: Since exponential is only defined for x≥0x \geq 0:

P(Z≤−264)=0P(Z \leq -\sqrt[4]{26}) = 0

Total:

P(Z4−17≥9)=e−4264≈0.000119P(Z^4 - 17 \geq 9) = e^{-4\sqrt[4]{26}} \approx 0.000119