Questions and Answers #10 Random Variables

Questions and Answers Sheet 10 Random Variables Question #1 What are the mean and standard deviation of a binomial probability distribution with n=190 and 𝑝 = 7 11 ? Answer: Explanation: For Binomial Probability Distribution mean 𝜇 = 𝑛𝑝 and standard deviation 𝜎 = √𝑛𝑝𝑞 7 Given n =190 and 𝑝 = 11 Compute the mean: 7 1330 10 𝜇 = 𝑛𝑝 = 190 ∗ = = 120 = 120.9 11 11 11 Compute the standard deviation 𝜎 𝜎 = √𝑛𝑝(1 − 𝑝) = √190 ∗ 𝜎 = √190 ∗ 7 11 7 11 4 2√1330 11 11 ∗( ) = ∗ (1 − 7 11 ) 𝜎 = 6.630 Question #2 What are the mean and standard deviation of a binomial probability distribution with n=144 and 𝑝 = 11 13 ? Answer: Explanation: The mean of a binomial distribution is simply: 𝜇 = 𝑛𝑝 Which in our case is 𝜇 = 144 ∗ 11 13 = 1584 13 ≅ 121.8 The standard deviation of the binomial distribution is given by: 𝜎 = √𝑛𝑝(1 − 𝑝) Which in our case is: 𝜎 = √144 ∗ 11 13 2 12 13 13 ∗( ) = √22 ≅ 4.330 Question #3 What are the mean and standard deviation of a binomial probability distribution with n=169 and 𝑝 = 1 13 ? Answer: Explanation: Mean and standard deviation of a binomial probability distribution are given by np and √𝑛𝑝𝑞 where q=1p. 1 1 12 In the given distribution, n=169 and 𝑝 = . Hence 𝑞 = 1 − 𝑝 = 1 − = . 13 Hence, mean is 169 ∗ or √12 = 3.464. 1 13 13 13 1 12 13 13 i.e. 13 and standard deviation is √169 ∗ ( ) ∗ ( ) Question #4 What are the mean and standard deviation of a binomial probability distribution with n=7 and 𝑝 = 19 21 ? Answer: 19 Mean of the distribution = 𝑛𝑝 = 7 × =6 21 Standard Deviation = √𝑛𝑝𝑞 = √7 × 19 21 × 1 3 2 21 =√ 38 63 = 0.7766 Question #5 What are the mean and standard deviation of a binomial probability distribution with n=777 and 𝑝 = 19 21 ? Answer: Explanation: 19 Mean = 𝑛𝑝 = 777 ∗ = 703 21 Variance = 𝑛𝑝(1 − 𝑝) = 777 ∗ 19 ∗ 21 2 21 = 66.9 Standard Deviation = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 8.18 Question #6 What are the mean and standard deviation of a binomial probability distribution with n=210 and 𝑝 = 12 21 ? Answer: Explanation: Binomial distribution is a frequency distribution of possible number of successful outcomes in given number of trials n, in each of which there is the probability of success is p (and q=1-p is probability of failure). The mean and standard deviation of a binomial distribution are given by np and √𝑛𝑝𝑞. Here 𝑛 = 210, 𝑝 = 12 21 and 𝑞 = 1 − Hence Mean is 210 × 12 21 1 21 × √22680 = 150.6 21 21 = 9 21 . = 120 Standard Deviation is √210 × = 12 12 21 × 9 21 = 7.17 Question #7 What are the mean and standard deviation of a binomial probability distribution with n=21 and 𝑝 = Answer: Explanation: For Binomial Probability Distribution 4 84 17 17 The Mean = 𝑛𝑝 = 21 ( ) = Standard Deviation 4 4 17 17 = √𝑛𝑝𝑞 = √𝑛𝑝(1 − 𝑝) = √21 ( ) (1 − ) = 1.94 4 17 ? Question #8 What are the mean and standard deviation of a binomial probability distribution with n=2 and 𝑝 = 4 17 ? Answer: The mean is 𝜇 = 𝑛 ∗ 𝑝 = 2 ∗ 𝜎 = √𝑛 ∗ 𝑝 ∗ (1 − 𝑝) = √2 ∗ 4 17 = 8 17 ≈ 0.4706 and the standard deviation is 4 13 104 ∗ =√ ≈ 0.5999 17 17 289 Explanation: The special formulas 𝜇 = 𝑛 ∗ 𝑝 and 𝜎 = √𝑛 ∗ 𝑝 ∗ (1 − 𝑝) only work for binomial random variables, where n is the number of independent trials with outcomes ''success'' or ''failure'', and p is the probability of ''success'' on each trial. The random variable X itself counts the number of successes in n trials. Question #9 What are the mean and standard deviation of a binomial probability distribution with n=15 and 𝑝 = 7 17 ? Answer: The mean is 𝜇 = 𝑛𝑝 = 105 17 ≈ 6,176 and the standard deviation is 𝜎 = √𝑛𝑝(1 − 𝑝) = 5√42 17 ≈ 1.906. Explanation: If X is a binomial random variable, counting the number of successes in n independent trials (where the only two outcomes are ''success'' and ''failure''), with constant probability of success p on each trial, the mean of X is 𝜇 = 𝑛𝑝 and the standard deviation is √𝑛𝑝(1 − 𝑝). In the present case, note that √𝑛𝑝(1 − 𝑝) = √15 ∗ 7 10 1050 √25 ∗ 42 5√42 ∗ =√ = = 17 17 289 17 √289 Question #10 What are the mean and standard deviation of a binomial probability distribution with n=5 and 𝑝 = Answer: Explanation: 10 Mean (n*p) of a Binomial probability distribution with n=5 and 𝑝 = is given by 5∗ 10 17 = 50 17 17 = 2.9412 Standard deviation of a Binomial probability distribution is given by √𝑛 ∗ 𝑝 ∗ 𝑞, where q=1-p. Hence, it is given by √5 ∗ 10 17 ∗ 7 17 i.e. √350 17 or 1.1 10 17 ?