Questions and Answers #12 Random Variables

Questions and Answers Sheet 12 Random Variables Question #1 For a population with a standard deviation of 𝜎 = 10 , a score of X = 44 corresponds to z = -0.50. What is the population mean? Answer: Explanation: The z-score is 𝑥−𝜇 𝑧= 𝜎 z=-0.5 x=44 𝜎 = 10 Therefore, 44−𝜇 −0.5 = 10 −5 = 44 − 𝜇 𝜇 = 44 + 5 = 49 Question #2 A population has a mean of 𝜇 = 100 and a standard deviation of 𝜎 = 10. If a single score is randomly selected from this population, how much distance, on average, should you find between the score and the population mean? Answer: Average distance of a value form mean is defined as the standard deviation. a) Average distance = 𝜎 = 10 b) Average distance of sample mean from population mean = standard error = 𝜎/√𝑛 = 10/2 = 5 Question #3 If a random sample of size 11 is drawn from a population distribution with standard deviation 𝜎 = 10 then what will be the variance of the sampling distribution of the sample mean? Answer: Explanation: GivenSample size = 11 Standard Deviation of the population distribution 𝜎 = 10 From the above information, we can get the standard deviation of the sampling distribution or standard error. 𝜎 10 𝑆𝐸 = = √𝑛 √11 10 2 Variance of the sampling distribution = ( ) = Variance of the sampling distribution = √𝑛 100 11 100 11 Question #4 What are the mean and standard deviation of a binomial probability distribution with n=5 and 𝑝 = 12 ? 17 Answer: Explanation: 12 60 Mean = 𝑛𝑝 = 5 ∗ = = 3.52 17 17 Standard deviation 12 12 = √𝑛𝑝𝑞 = √𝑛𝑝(1 − 𝑝) = √5 ( ) (1 − ) = 1.01 17 17 Question #5 What are the mean and standard deviation of a binomial probability distribution with n=8 and 𝑝 = 12 ? 19 Answer: Explanation: For a binomial distribution Mean 𝜇 = 𝑛∗𝑝 In this case: 12 𝜇 = 8 ∗ ≈ 5.05 19 Standard Deviation 𝜎 = √𝑛 ∗ 𝑝 ∗ (1 − 𝑝) In this case 𝜎 = √8 ∗ 12 19 ∗ 7 19 ≈ 1.36 Question #6 What are the mean and standard deviation of a binomial probability distribution with n=16 and 𝑝 = 12 47 ? Answer: Explanation: The mean and standard deviation of the Binomial distribution are given by: 𝜇 = 𝑛𝑝 𝜎 = √𝑛𝑝(1 − 𝑝) Note that the standard deviation is simply the square root of the variance. Plugging in the numbers we get: 12 192 47 47 𝜇 = 16 ∗ ( ) = 192 and 𝜎 = √ 47 ≅ 4.09 ∗ (1 − 12 47 )= √6720 47 ≅ 1.74 Question #7 What are the mean and standard deviation of a binomial probability distribution with n=4 and 𝑝 = Answer: Explanation: Given n=4 15 32 ? 𝑝= 15 32 𝑞 =1−𝑝 =1− 15 = 17 32 32 15 7 Mean = 𝑛𝑝 = 4 × 32 𝑆𝐷 = √𝑛𝑝𝑞 = √4 × =1 15 32 × 8 17 32 =√ 1020 1024 =√ 255 256 = 0.998 Result: 7 Mean = 1 8 SD = 0.998 Question #8 What are the mean and standard deviation of a binomial probability distribution with n=9 and 𝑝 = 11 32 ? Answer: mean =np 𝑠. 𝑑. = √𝑛𝑝𝑞 Explanation: Using the formula for mean and s.d. above ... 11 99 mean = 9 × = ≈ 3.09 32 s.d. = √9 × 11 32 32 × (1 − 11 32 ) ≈ 1.425 Question #9 What are the mean and standard deviation of a binomial probability distribution with n=12 and 𝑝 = 31 32 ? Answer: Explanation: The mean and standard deviation of a binomial probability distribution is given by np and √𝑛𝑝𝑞, where q-1-p. With n=12 and 𝑝 = Mean = 12 ∗ 31 32 31 32 ,𝑞 = 1 − 31 32 or 𝑞 = 1 32 , = 11.625 and 31 1 1 19.2 32 32 32 32 Standard Deviation = √12 ∗ ( ) ∗ ( ) = ( ) ∗ √372 = = 0.6 Question #10 What are the mean and standard deviation of a binomial probability distribution with n=16 and p = 21 32 ? Answer: Explanation: There are standard formulae for the mean and standard deviation of a binomial probability distribution. If you are going to do any significant work with binomial probability distributions you should memorize these: 𝜇 = 𝑛∗𝑝 𝜎 = √𝑛 ∗ 𝑝 ∗ (1 − 𝑝) mean: 𝜇 = 10.5 standard deviation: 𝜎 = 1.9