Questions and Answers #13 Random Variables

Questions and Answers Sheet 13 Random Variables Question #1 What are the mean and standard deviation of a binomial probability distribution with n=17 and 𝑝 = 18 32 ? Answer: Explanation: Unless you have to calculate everything each time, we could simply use known formulas. In binomial distribution the mean is given by np and variance by np(1-p). Since in our case n=17 and 𝑝 = (1 − 9 16 )= 1071 256 18 32 = 9 16 the mean is 17 ∗ 9 16 = 153 16 = 9.5 and variance is 17 ∗ . Standard deviation is the square root of variance so we have √ 1071 256 =3 √119 16 9 16 ∗ ≈ 2.0454. Question #2 What are the mean and standard deviation of a binomial probability distribution with n = 88 and 𝑝 = 24 32 ? Answer: Explanation: for a Binomial distribution 𝑋 ∼ 𝐵(𝑛, 𝑝) in this case 24 𝑋 ∼ 𝐵 (88, ) 32 3 or 𝑋 ∼ 𝐵 (88, ) 4 mean E(X)=np in this case 3 𝐸(𝑋) = 88 × = 66 4 standard deviation = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 sd = √𝑛𝑝(1 − 𝑝) sd = √66 × 1 4 sd = √22 Question #3 What are the mean and standard deviation of a binomial probability distribution with n=18 and 𝑝 = 2 32 ? Answer: Explanation: A binomial random variable X counts the number of successes in n trials, when each trial is independent from the others with a constant probability p of success. The mean of X is 𝜇 = 𝑛 ∗ 𝑝 and the standard deviation of X is 𝜎 = √𝑛 ∗ 𝑝 ∗ (1 − 𝑝). The mean is 𝜇 = 𝑛 ∗ 𝑝 = 18 ∗ 𝜎 = √𝑛 ∗ 𝑝 ∗ (1 − 𝑝) = √ 1 16 = 9 8 and the standard deviation is 135 3√15 3√30 = = ≈ 1.027 128 16 8√2 Question #4 What are the mean and standard deviation of a binomial probability distribution with n=5 and 𝑝 = 1 32 ? Answer: For the Binomial distribution ... mean =np variance =npq Explanation: 1 5 mean = 𝑛𝑝 = 5 × = 32 variance = 𝑛𝑝𝑞 = 5 × 32 1 1 32 32 × (1 − ) = 155/1024 155 standard deviation = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = √ ≈ 0.389 1024 Question #5 What are the mean and standard deviation of a probability density function given by 𝑃𝑟(𝑋 = 𝑘) = 3𝑘 𝑒 −3 𝑘! for 𝑘 ∈ {0,1,2 … ∞}? Answer: We can recognize that is the PMF of Poisson distribution with 𝜆 = 3 The pmf of Poisson distribution is 𝜆𝑘 𝑒 −𝜆 𝑘! And the mean is 𝜇 = 𝜆 = 3 , and the standard deviation is 𝜎 = √𝜆 = √3 = 1.73 Mean is 3, standard deviation is 1.732 Question #6 What are the mean and standard deviation of a probability density function given by 𝑃𝑟(𝑋 = 𝑘) = 12𝑘 𝑒 −12 𝑘! for 𝑘 ∈ {0,1,2 … ∞}? Answer: Explanation: Probability of events for a Poisson distribution An event can occur 0, 1, 2, … times in an interval. The average number of events in an interval is designated 𝜆 . 𝜆 is the event rate, also called the rate parameter. The probability of observing k events in an interval is given by the equation P(k events in interval) = 𝜆𝑘 𝑒 −𝜆 𝑘! The mean is equal to the variance mean = 𝜆 = 𝜎 2 and standard deviation 𝜎 = √𝜆 mean 𝜆 = 12 standard deviation √𝜆 = √12 = 3.464 Question #7 What are the mean and standard deviation of a probability density function given by 𝑃𝑟(𝑋 = 𝑘) = for 𝑘 ∈ {0,1,2 … ∞} ? 80𝑘 𝑒 −80 𝑘! Answer: Explanation: 𝑃(𝑥 = 𝑘) = 80𝑘 𝑒 −80 𝑘! , 𝑘 ∈ {0,1,2,3 … ∞} is a Poisson distribution, the standard formula being: 𝑃(𝑥 = 𝑘) = 𝜆𝑘 𝑒 −𝜆 𝑘! , 𝑘 ∈ {0,1,2,3 … ∞} the mean and variance are given by E(X) = 𝜆 Var(X) = 𝜆 the proof of these results can be found from a search on the web. the sd is the square root of the variance hence the results E(X)=80, Var(X)=80 𝑠𝑑 = √80 Question #8 What are the mean and standard deviation of a probability density function given by 𝑃𝑟(𝑋 = 𝑘) = 14𝑘 𝑒 −14 𝑘! for 𝑘 ∈ {0,1,2 … ∞} ? Answer: Explanation: The mean and variance of the Poisson distribution is 𝜆 So for this problem ... 𝜇 = 𝜎 2 = 14 Standard Deviation = √𝜆 = √14 This is the Poisson distribution with 𝜆 = 14 Question #9 What are the mean and standard deviation of a probability density function given by 𝑃𝑟(𝑋 = 𝑘) = for 𝑘 ∈ {0,1,2 … ∞} ? Answer: Explanation: This is a Poisson distribution. the standard eqn is: 𝑋 ∼ 𝑃(𝜆) ⇒ 𝑃(𝑋 = 𝑘) = 𝜆𝑘 𝑒 −𝜆 𝑘! For 𝑘 ∈ {0,1,2,3,4. . ∞} where mean 𝐸(𝑋) = 𝜆; Variance 𝑉𝑎𝑟(𝑋) = 𝜆. in this case 𝑃(𝑋 = 𝑘) = 10𝑘 𝑒 −10 𝑘! For 𝑘 ∈ {0,1,2,3,4. . ∞} ∴ 𝐸(𝑋) = 10 Var(X)=10 ∴ 𝑠𝑑 = √10 Question #10 What are the mean and standard deviation of a probability density function given by Pr(X=k)=\frac{9^{k}e^{-9}}{k!} for k\in\left\{0,1,2...\infty\right\} ? Answer: 10𝑘 𝑒 −10 𝑘! Explanation: this pdf is a Poisson distribution X\sim Po(\lambda) P(X=k)=\frac{\lambda^{k}e^{-k}}{k!}, k\in \left\{0,1,2,3,4, ... \infty \right\} In such cases the mean (E(X)) and variance (Var(X)) are given by E(X)=\lambda, Var(X)=\lambda In this case we have P(X=k)=\frac{9^{k}e^{-9}}{k!}, k\in \left\{0,1,2,3,4, ... \infty \right\} so E(X)=9 Var(X)=9 sd=\sqrt{Var}=\sqrt{9}=3