University:
Rice University
Course:
STAT 310 | Probability and Statistics
Academic year:

2021
Views:

14

Pages:

2
Author:

MathMonk

Questions and Answers Sheet 14 Random Variables Question #1 What are the mean and standard deviation of a probability density function given by 𝑃𝑟(𝑋 = 𝑘) = 18𝑘 𝑒 −18 𝑘! for 𝑘 ∈ {0,1,2 … ∞} ? Answer: Explanation: The mean and variance of the Poisson distribution are both equal to 𝜆 𝜇 = 𝜆 = 18 𝜎 = √𝜆 = √18 = 3√2 ≈ 4.243 Question #2 What are the mean and standard deviation of a probability density function given by 𝑃𝑟(𝑋 = 𝑘) = 24𝑘 𝑒 −24 𝑘! for 𝑘 ∈ {0,1,2 … ∞} ? Answer: Explanation: 𝑃(𝑋 = 𝑘) = 24𝑘 𝑒 −24 𝑘! for 𝑘 ∈ {0,1,2 … ∞} is a Poisson distribution which is of teh form 𝑃(𝑋 = 𝑘) = 𝜆𝑘 𝑒 −𝜆 𝑘! for 𝑘 ∈ {0,1,2 … ∞} This means that E(X) = 𝜆 and Var(X) = 𝜆 for the given distribution E(X)=24 Var(X)=24 𝑠𝑑 = √𝑉𝑎𝑟(𝑋) = √24 = 2√6 Question #3 What are the mean and standard deviation of the probability density function given by 𝑝(𝑥) 𝑘 = sin 𝑥 for 𝑥 ∈ [0, 𝜋], in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1? Answer: Explanation: The recipe is: Step 1 Determine the probability distribution function pdf, p(x) such that: 𝑥 𝑃(𝑥) = ∫𝑥 2 𝑝(𝑥)𝑑𝑥 = 1 over [𝑥1 , 𝑥2 ] 𝜋 1 1 = ∫0 𝑘 sin(𝑥) 𝑑𝑥 = 2𝑘; 𝑘 = 1 1 2 so your pdf is 𝑝(𝑥) = sin 𝑥 ; [0, 𝜋] 2 Step 2 Determine the mean \mu and standard deviation 𝜎 𝜋 1 a) Mean: 𝜇 = ∫0 𝑥 sin 𝑥 𝑑𝑥 = 2 1 sin 𝑥−𝑥 cos 𝑥 2 𝜋 b) var: 𝜎 2 = ∫0 (𝑥 − 𝜇)2 sin 𝑥 𝑑𝑥 2 4(2𝑥−𝜋) sin 𝑥−((2𝑥−𝜋)2 −8) cos 𝑥 𝜎2 = 2 1 = = 𝜋 2 𝜋2 −8 4 𝜎 = √𝜋 2 − 8 2 Question #4 A population has a mean of 24.8 and a standard deviation of 4.2. If 𝑥𝑖 = 19, what is the z-value? Answer: Explanation: Given Mean 𝑥 = 24.8 Standard Deviation =4.2 𝑥𝑖 = 19 𝑧= 𝑧= 𝑥−𝑥 𝜎 19−24.8 4.2 = −5.8 4.2 = −1.38 z=-1.38 Question #5 The sum of all X's is 3780. The sum of all 𝑋 2 is 80000. The sample N = 180. How can I calculate the variance? Answer: Explanation: ∑ 𝑥 = 3780 ∑ 𝑥 2 = 80000 n=180 𝑥= ∑𝑥 𝑛 𝑉(𝑥) = = = 3780 180 = 21 2 ∑ 𝑥 2 −𝑥 180 80000−212 79559 180 = 180 = 442