Random variables #7 - Q&A

Question 1 What is the difference between variance and variability? Answer 1 Variability, which measures how much the data varies, is defined as "lack of consistency." Measures of variability include variance, standard deviation, interquartile range, and range. The average squared deviation of a random variable from its mean is known as variance. Variance of 𝑋 X is defined as Var ( 𝑋 ) = 𝐸 [ ( 𝑋 − 𝜇 ) 2 ] Var(X)=E[(X−μ) 2 ]. Answer 2 The short answer is no, variation ≠ variance, as stated by others in the comments section. Spread, dispersion, scatter, and variability are all alternatives for the noun "variation." It's just a way of talking about the behavior of the data in a general sense as either having a lot of density over a narrow interval (generally near the mean, but not necessarily if the distribution is skewed) or spread out over a wide range. Variance is a particular measure of variability, but others exist (and several are enumerated in the linked article). Question 2 How do you find the variance of the data 2, 4, 6, 8, 10? Answer 1 Explanation: Variance ( 𝜎 2 ) (σ 2 ) is the average of squared difference from the mean. Mean is 𝑀 = 2 + 4 + 6 + 8 + 10 5 = 30 5 = 6 M= 5 2+4+6+8+10 ​ = 5 30 ​ =6. Variance is 𝜎 2 = ( 2 − 6 ) 2 + ( 4 − 6 ) 2 + ( 6 − 6 ) 2 + ( 8 − 6 ) 2 + ( 10 − 6 ) 2 5 σ 2 = 5 (2−6) 2 +(4−6) 2 +(6−6) 2 +(8−6) 2 +(10−6) 2 ​ : = 16 + 4 + 0 + 4 + 16 5 = 40 5 = 8. = 5 16+4+0+4+16 ​ = 5 40 ​ =8. So the variance is 8 [Ans]. Answer 2 Mean, 𝑀 = 2 + 4 + 6 + 8 + 10 5 = 6 M= 5 2+4+6+8+10 ​ =6. Variance: 𝜎 2 = 1 5 ∑ 𝑖 = 1 𝑛 ( 𝑥 𝑖 − 𝑥 ˉ ) 2 σ 2 = 5 1 ​ i=1 ∑ n ​ (x i ​ − x ˉ ) 2 = 1 5 { ( 2 − 6 ) 2 + ( 4 − 6 ) 2 + ( 6 − 6 ) 2 + ( 8 − 6 ) 2 + ( 10 − 6 ) 2 } . = 5 1 ​ {(2−6) 2 +(4−6) 2 +(6−6) 2 +(8−6) 2 +(10−6) 2 }. = 1 5 { 16 + 4 + 0 + 4 + 16 } = 40 5 = 8. = 5 1 ​ {16+4+0+4+16}= 5 40 ​ =8. Question 3 What does the variance of a random variable measure? What is the standard deviation of a continuous random variable 𝑋 X? Answer 1 Step 1 A random variable's variance quantifies how widely distributed a set of (random) integers are relative to their mean. Step 2 A random variable with an unlimited number of possible values is said to be continuous. Any value between the ranges of values is possible for the continuous random variable. The mean deviation of the random variable from the mean value is known as the standard deviation of any variable. It establishes how the data values are distributed around the mean. A variable's distribution will be closer to the mean value if the standard deviation is less. A variable's distribution will be further from the mean value if the standard deviation is higher. The following is the definition of the formula for the continuous random variable 𝑋 X's standard deviation: 𝜎 = ∫ − ∞ ∞ ( 𝑥 − 𝜇 ) 2 𝑓 ( 𝑥 ) 𝑑 𝑥 . σ= ∫ −∞ ∞ ​ (x−μ) 2 f(x)dx ​ . Question 4 Consider a set of 20 independent and identically distributed uniform random variables in the interval ( 0 , 1 ) (0,1). What is the expected value and variance of the sum, 𝑆 S, of these random variables? Select one: (a) 𝐸 ( 𝑆 ) = 10 ,   Var ( 𝑆 ) = 1.66667 E(S)=10,Var(S)=1.66667 (b) 𝐸 ( 𝑆 ) = 200 ,   Var ( 𝑆 ) = 16.6667 E(S)=200,Var(S)=16.6667 (c) 𝐸 ( 𝑆 ) = 100 ,   Var ( 𝑆 ) = 200 E(S)=100,Var(S)=200 (d) 𝐸 ( 𝑆 ) = 10 ,   Var ( 𝑆 ) = 12 E(S)=10,Var(S)=12. Answer 1 Step 1 Uniform distribution: A continuous random variable 𝑋 X is said to follow Uniform distribution if the probability density function of 𝑋 X is: 𝑓 ( 𝑥 ) = 1 𝑏 − 𝑎 , where  𝑎  is lower limit and  𝑏  is the upper limit of  𝑥 . f(x)= b−a 1 ​ ,where a is lower limit and b is the upper limit of x. The mean of the Uniform distribution is 𝜇 = 𝑎 + 𝑏 2 μ= 2 a+b ​ . The variance of the Uniform distribution is 𝜎 2 = ( 𝑏 − 𝑎 ) 2 12 σ 2 = 12 (b−a) 2 ​ . Step 2 In this case, the i.i.d random variables 𝑋 1 , 𝑋 2 , … , 𝑋 20 X 1 ​ ,X 2 ​ ,…,X 20 ​ follow uniform distribution over the interval ( 0 , 1 ) (0,1). Consider a new random variable 𝑆 = 𝑋 1 + 𝑋 2 + … + 𝑋 20 S=X 1 ​ +X 2 ​ +…+X 20 ​ . Hence, the mean of 𝑆 S is: 𝐸 ( 𝑆 ) = 𝐸 ( 𝑋 1 ) + 𝐸 ( 𝑋 2 ) + … + 𝐸 ( 𝑋 20 ) . E(S)=E(X 1 ​ )+E(X 2 ​ )+…+E(X 20 ​ ). = 20 𝐸 ( 𝑋 ) ( as independent and identically distributed ) . =20E(X)(as independent and identically distributed). = 20 ( 0 + 1 2 ) = 10. =20( 2 0+1 ​ )=10. Hence, the variance of 𝑆 S is: 𝑉 ( 𝑆 ) = 𝑉 ( 𝑋 1 ) + 𝑉 ( 𝑋 2 ) + … + 𝑉 ( 𝑋 20 ) . V(S)=V(X 1 ​ )+V(X 2 ​ )+…+V(X 20 ​ ). = 20 𝑉 ( 𝑋 ) ( as independent and identically distributed ) . =20V(X)(as independent and identically distributed). = 20 ( ( 1 − 0 ) 2 12 ) = 1.66667. =20( 12 (1−0) 2 ​ )=1.66667. Thus, the correct option is: (a) 𝐸 ( 𝑆 ) = 10 ,   Var ( 𝑆 ) = 1.66667 E(S)=10,Var(S)=1.66667