Random variables #5 - Q&A

Question 1 What is a discrete probability distribution? What two conditions determine a probability distribution? Answer 1 Solution: Discrete distribution is a probability distribution where sample space is countable. The random variable in such a case can only take point values. Example: Number of people, or number of coins, etc. The two conditions are: Sum of all probabilities should be equal to 1. Each probability must be ≥ 0 ≥0. Answer 2 A discrete pdf is a pdf of a discrete variable like the number of eyes thrown by a dice. The sum of all probabilities is 1, and the probabilities are ≥ 0 ≥0. Solution: A discrete variable has a finite number of distinct values, similar to dice’s tossed eyes: 1, 2, 3, 4, 5, or 6. The discrete pdf for a dice throw is: 𝑃 [ 𝑥 = 𝑘 ] = 1 / 6 , 𝑘 = 1 , 2 , … , 5 , 6. P[x=k]=1/6,k=1,2,…,5,6. The pdf for the amount of rainfall in a month is not discrete since the values are continuous and there is no upper bound. Question 2 A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, what is the probability that a head will show between 30 and 36 times inclusive? Answer 1 0.505 Explanation: 𝑛 = 60 𝑝 = 1 2 , 𝑚 = 𝑛 𝑝 = 30 , n=60p= 2 1 ​ ,m=np=30, and 𝑠 = 𝑛 𝑝 ( 1 − 𝑝 ) = 15 . s= np(1−p) ​ = 15 ​ . 𝑧 2 = 36.5 − 30 15 = 1.678. z 2 ​ = 15 ​ 36.5−30 ​ =1.678. 𝑧 1 = 29.5 − 30 15 = − 0.129. z 1 ​ = 15 ​ 29.5−30 ​ =−0.129. We search 𝑧 1 z 1 ​ and 𝑧 2 z 2 ​ in a table for z-values and get: 0.953 − 0.448 = 0.505. 0.953−0.448=0.505. Question 3 What is Chebyshev’s inequality? Answer 1 Chebyshev’s inequality says that at least 1 − 1 𝑘 2 1− k 2 1 ​ of data from a sample must fall within 𝑘 k standard deviations from the mean, where 𝑘 k is any positive real number greater than one. Explanation: Let’s play with a few values of 𝑘 k: 𝑘 = 2 k=2, we have 1 − 1 2 2 = 1 − 1 4 = 75 % 1− 2 2 1 ​ =1− 4 1 ​ =75%. So Chebyshev’s would tell us that 75% of the data values of any distribution must be within two standard deviations of the mean. 𝑘 = 3 k=3, we have 1 − 1 3 2 = 1 − 1 9 = 89 % 1− 3 2 1 ​ =1− 9 1 ​ =89%. This time we have 89% of the data values within three standard deviations of the mean. 𝑘 = 4 k=4, we have 1 − 1 4 2 = 1 − 1 16 = 93.75 % 1− 4 2 1 ​ =1− 16 1 ​ =93.75%. Now we have 93.75% of the data within four standard deviations of the mean. This is consistent with saying that in a Normal distribution, 68% of the data is one standard deviation from the mean, 95% is two standard deviations from the mean, and approximately 99% is within three standard deviations from the mean. The difference is Chebyshev’s theorem extends this principle to any distribution. Answer 2 Chebyshev’s inequality is a probability theory that guarantees that within a specified range or distance from the mean, for a large range of probability distributions, no more than a specific fraction of values will be present. Formula: 1 − 1 𝑘 2 , 1− k 2 1 ​ , where 1 𝑘 2 k 2 1 ​ of a distribution’s values can be no more than or equal to 𝑘 k standard deviations away from the mean of the distribution. Question 4 What are the mean and standard deviation of a binomial probability distribution with 𝑛 = 12 n=12 and 𝑝 = 7 60 p= 60 7 ​ ? Answer 1 Explanation: The mean and standard deviation of a binomial probability distribution are 𝑛 𝑝 np and 𝑛 𝑝 𝑞 npq ​ , respectively, where 𝑛 n is the number of trials and 𝑝 p is the probability of success of the event. Here 𝑞 = 1 − 𝑝 q=1−p, and is the probability of failure of the event. With  𝑛 = 12  and  𝑝 = 7 60 , With n=12 and p= 60 7 ​ , Mean  = 12 × 7 60 = 84 60 = 1.4 , Mean =12× 60 7 ​ = 60 84 ​ =1.4, and Standard Deviation  = 𝑛 𝑝 𝑞 = 12 × 7 60 × 53 60 . Standard Deviation = npq ​ = 12× 60 7 ​ × 60 53 ​ ​ . = 4452 3600 = 66.7233 60 = 1.112. = 3600 4452 ​ ​ = 60 66.7233 ​ =1.112. Question 5 What are the mean and standard deviation of a binomial probability distribution with 𝑛 = 2 n=2 and 𝑝 = 4 17 p= 17 4 ​ ? Answer 1 The mean is: 𝜇 = 𝑛 ⋅ 𝑝 = 2 ⋅ 4 17 = 8 17 ≈ 0.4706. μ=n⋅p=2⋅ 17 4 ​ = 17 8 ​ ≈0.4706. The standard deviation is: 𝜎 = 𝑛 ⋅ 𝑝 ⋅ ( 1 − 𝑝 ) = 2 ⋅ 4 17 ⋅ 13 17 . σ= n⋅p⋅(1−p) ​ = 2⋅ 17 4 ​ ⋅ 17 13 ​ ​ . 𝜎 = 104 289 ≈ 0.5999. σ= 289 104 ​ ​ ≈0.5999. Explanation: The formulas 𝜇 = 𝑛 ⋅ 𝑝 μ=n⋅p and 𝜎 = 𝑛 ⋅ 𝑝 ⋅ ( 1 − 𝑝 ) σ= n⋅p⋅(1−p) ​ only work for binomial random variables, where 𝑛 n is the number of independent trials with outcomes success or failure, and 𝑝 p is the probability of success on each trial.