=====Mathematics 340 - Fall 2015=====

====Homework====

**Exercises 4.12**: 21, 28, 60, 65, 76(a)(b)


Also:

Suppose 1 out of every 10,000 people have a certain disease. Let $X$ be the number of people in a sample of 20,000 that have disease.
  - What is $\text{E}(X)$?
  - Find, exactly, $P(X = 0)$, $P(X = 1)$, $P(X = 2)$, and $P(X > 3)$
  - Find the appropriate approximations for the above probabilities.

Answers:

  - $E(X) = 200000 \cdot \dfrac{1}{10000} = 2$
  - $P(X = 0) = (0.9999)^{20000} = 0.135322$, $P(X = 1) = 20000 \cdot (0.0001) \cdot (0.9999)^{19999} = 0.270671$, $P(X = 2) = \binom{20000}{2} \cdot (0.0001)^2 \cdot (0.9999)^{19998} = 0.270684$, $P(X > 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) = 0.1428675$
  - Using the Poisson approximation for binomial probabilities, $P(X = 0) \approx e^{-2} = 0.135335$, $P(X = 1) \approx 2e^{-2} = 0.270671$, $P(X = 2) \approx \frac{2^2 \cdot e^{-2}}{2} = 0.270671$, $P(X > 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) \approx 0.1428765$