Random variables #6 - Q&A

Question 1 What are the mean and standard deviation of a binomial probability distribution with 𝑛 = 17 n=17 and 𝑝 = 18 32 p= 32 18 ​ ? Answer 1 Explanation: Unless you have to calculate everything each time, we could simply use known formulas. In a binomial distribution, the mean is given by 𝑛 𝑝 np and variance by 𝑛 𝑝 ( 1 − 𝑝 ) np(1−p). Since in our case 𝑛 = 17 n=17 and 𝑝 = 18 32 = 9 16 p= 32 18 ​ = 16 9 ​ : The mean is  17 ⋅ 9 16 = 9.56 , The mean is 17⋅ 16 9 ​ =9.56, and Variance is  17 ⋅ 9 16 ⋅ ( 1 − 9 16 ) = 1071 256 . Variance is 17⋅ 16 9 ​ ⋅(1− 16 9 ​ )= 256 1071 ​ . Standard deviation is the square root of variance, so we have: 𝜎 = 1071 256 = 3 119 16 ≈ 2.0454. σ= 256 1071 ​ ​ = 16 3 119 ​ ​ ≈2.0454. Answer 2 Solution: The mean of the binomial distribution is interpreted as the mean number of successes for the distribution. To find the mean, use the formula: 𝜇 = 𝑛 ⋅ 𝑝 . μ=n⋅p. Substituting values for this problem, we have: 𝜇 = 17 ⋅ 18 32 = 9.5625. μ=17⋅ 32 18 ​ =9.5625. The standard deviation of the binomial distribution is interpreted as the standard deviation of the number of successes for the distribution. To find the standard deviation, use the formula: 𝜎 = 𝑛 ⋅ 𝑝 ⋅ ( 1 − 𝑝 ) . σ= n⋅p⋅(1−p) ​ . Substituting values, we have: 𝜎 = 17 ⋅ 18 32 ⋅ ( 1 − 18 32 ) . σ= 17⋅ 32 18 ​ ⋅(1− 32 18 ​ ) ​ . Evaluating: 𝜎 = 4.1888 ≈ 2.0454. σ= 4.1888 ​ ≈2.0454. Question 2 Let 𝑋 X and 𝑌 Y be Bernoulli random variables. Let 𝑍 = 𝑋 𝑌 Z=XY. a) Show that 𝑍 Z is a Bernoulli random variable. b) Show that if 𝑋 X and 𝑌 Y are independent, then 𝑃 𝑍 = 𝑃 𝑋 ⋅ 𝑃 𝑌 . P Z ​ =P X ​ ⋅P Y ​ . Answer 1 Step 1 Given: 𝑋 X and 𝑌 Y are Bernoulli random variables. Bernoulli random variables are discrete random variables that have outcomes 0 or 1. a) 𝑍 = 𝑋 𝑌 Z=XY: As the possible values of 𝑋 X and 𝑌 Y are 0 or 1, the possible values of 𝑍 = 𝑋 𝑌 Z=XY, for all values of 𝑋 X and 𝑌 Y, are also 0 or 1. Hence, 𝑍 = 𝑋 𝑌 Z=XY is also a Bernoulli random variable. Step 2 b) Given 𝑋 X and 𝑌 Y are independent events: 𝑃 ( 𝑋  and  𝑌 ) = 𝑃 ( 𝑋 ) ⋅ 𝑃 ( 𝑌 ) . P(X and Y)=P(X)⋅P(Y). Let us consider 𝑃 ( 𝑍 = 1 ) P(Z=1): 𝑃 ( 𝑍 = 1 ) = 𝑃 ( 𝑋 = 1  and  𝑌 = 1 ) . P(Z=1)=P(X=1 and Y=1). As both are independent: 𝑃 ( 𝑍 = 1 ) = 𝑃 ( 𝑋 = 1 ) ⋅ 𝑃 ( 𝑌 = 1 ) . P(Z=1)=P(X=1)⋅P(Y=1). Thus: 𝑃 𝑍 = 𝑃 𝑋 ⋅ 𝑃 𝑌 . P Z ​ =P X ​ ⋅P Y ​ . Hence proved. Question 3 To explain: Linearity of expectation of random variables. Answer 1 Given: 𝑛 n random variables 𝑋 1 , 𝑋 2 , … , 𝑋 𝑛 X 1 ​ ,X 2 ​ ,…,X n ​ . Formula used: 𝐸 ( 𝑋 1 + 𝑋 2 + … + 𝑋 𝑛 ) = 𝐸 ( 𝑋 1 ) + 𝐸 ( 𝑋 2 ) + … + 𝐸 ( 𝑋 𝑛 ) . E(X 1 ​ +X 2 ​ +…+X n ​ )=E(X 1 ​ )+E(X 2 ​ )+…+E(X n ​ ). Linearity means additivity over the variables. Thus, this expresses the fact that expectation is additive over the random variables. In other words, expectation is a linear function. Question 4 How do you find the standard deviation of a set of numbers? Answer 1 Population standard deviation: 𝜎 = ( 𝑥 1 − 𝑥 ˉ ) 2 + ( 𝑥 2 − 𝑥 ˉ ) 2 + … + ( 𝑥 𝑛 − 𝑥 ˉ ) 2 𝑛 σ= n (x 1 ​ − x ˉ ) 2 +(x 2 ​ − x ˉ ) 2 +…+(x n ​ − x ˉ ) 2 ​ ​ Sample standard deviation: 𝑠 = ( 𝑥 1 − 𝑥 ˉ ) 2 + ( 𝑥 2 − 𝑥 ˉ ) 2 + … + ( 𝑥 𝑛 − 𝑥 ˉ ) 2 𝑛 − 1 s= n−1 (x 1 ​ − x ˉ ) 2 +(x 2 ​ − x ˉ ) 2 +…+(x n ​ − x ˉ ) 2 ​ ​ Steps: Find the mean of the set of numbers: 𝑥 ˉ = 𝑥 1 + 𝑥 2 + … + 𝑥 𝑛 𝑛 , x ˉ = n x 1 ​ +x 2 ​ +…+x n ​ ​ , where 𝑛 n = the number of numbers in the set. Subtract the mean from each number in your sample, square the difference, and add: ( 𝑥 1 − 𝑥 ˉ ) 2 + ( 𝑥 2 − 𝑥 ˉ ) 2 + … + ( 𝑥 𝑛 − 𝑥 ˉ ) 2 . (x 1 ​ − x ˉ ) 2 +(x 2 ​ − x ˉ ) 2 +…+(x n ​ − x ˉ ) 2 . Divide by 𝑛 n (for population) or 𝑛 − 1 n−1 (for sample). Take the square root. Answer 2 Calculate the Mean (the arithmetic average of the numbers). For each value, take the Mean out and square the result. Calculate the mean of those squared differences. We are done after taking that square root. Question 5 If you add two independent random variables, what is the standard deviation of the combined distribution if the standard deviations of the two original distributions were, for example, 7 and 5? Answer 1 You cannot just add the standard deviations. Instead, you add the variances. Those are built up from the squared differences between every individual value from the mean. Standard deviation is defined as the square root of the variance. The other way around, variance is the square of SD. Steps: You square the individual SDs to get the variances. Then you add those together to get the total variance. Then you take the square root to get the total SD. Your case: Total variance = 7 2 + 5 2 = 49 + 25 = 74. 7 2 +5 2 =49+25=74. Total SD = 74 ≈ 8.60. 74 ​ ≈8.60.