Random variables #3 - Q&A

Question 1 A random variable 𝑋 X has the discrete uniform distribution 𝑓 ( 𝑥 ; 𝑘 ) = 1 𝑘 ,   𝑥 = 1 , 2 , … , 𝑘 f(x;k)= k 1 ​ ,x=1,2,…,k 𝑓 ( 𝑥 ; 𝑘 ) = 0 ,   elsewhere . f(x;k)=0,elsewhere. Show that the moment-generating function of 𝑋 X is 𝑀 𝑋 ( 𝑡 ) = 𝑒 𝑡 ( 1 − 𝑒 𝑘 𝑡 ) 𝑘 ( 1 − 𝑒 𝑡 ) . M X ​ (t)= k(1−e t ) e t (1−e kt ) ​ . Answer 1 By using the definition of the moment-generating function of a discrete random variable 𝑋 X, we get: 𝑀 𝑋 ( 𝑡 ) = 𝐸 ( 𝑒 𝑡 𝑋 ) = ∑ 𝑥 = 1 𝑘 𝑒 𝑡 𝑥 𝑓 ( 𝑥 ) = ∑ 𝑥 = 1 𝑘 𝑒 𝑡 𝑥 ⋅ 1 𝑘 . M X ​ (t)=E(e tX )= x=1 ∑ k ​ e tx f(x)= x=1 ∑ k ​ e tx ⋅ k 1 ​ . Simplify: 𝑀 𝑋 ( 𝑡 ) = 1 𝑘 ∑ 𝑥 = 1 𝑘 𝑒 𝑡 𝑥 . M X ​ (t)= k 1 ​ x=1 ∑ k ​ e tx . Use the formula for a geometric series: 𝑀 𝑋 ( 𝑡 ) = 1 𝑘 ⋅ 𝑒 𝑡 ( 1 − 𝑒 𝑘 𝑡 ) 1 − 𝑒 𝑡 . M X ​ (t)= k 1 ​ ⋅ 1−e t e t (1−e kt ) ​ . Thus: 𝑀 𝑋 ( 𝑡 ) = 𝑒 𝑡 ( 1 − 𝑒 𝑘 𝑡 ) 𝑘 ( 1 − 𝑒 𝑡 ) . M X ​ (t)= k(1−e t ) e t (1−e kt ) ​ . Question 2 You are concerned about the effects of texting on driving performance. Design a study to test the response time of drivers while texting and while driving only. How many seconds does it take for a driver to respond when a leading car hits the brakes? a. Describe the explanatory and response variables in the study. b. What are the treatments? c. What should you consider when selecting participants? d. Your research partner wants to divide participants randomly into two groups: one to drive without distraction and one to text and drive simultaneously. Is this a good idea? Why or why not? e. Identify any lurking variables that could interfere with this study. f. How can blinding be used in this study? Answer 1 a. The explanatory variable is whether the test subject is texting and driving or simply driving. The response variable is the amount of time it takes the driver to respond to the car in front of them hitting the brakes. b. There are two treatments: requiring a driver to text while driving and requiring a driver to not text while driving. c. You should consider the age of the driver, the level of skill of the driver, and the previous driving history of the driver. d. Dividing the participants randomly is a good idea, as it controls for lurking variables. However, this experiment must be done in controlled areas, not simply out on the road, as the results could be dangerous to other drivers. e. Some lurking variables could include the level of skill of the driver, the frequency with which the driver normally texts and drives, and the previous driving history of the driver. f. Blinding can be used by ensuring participants do not know which aspect of their performance (texting or not texting) is being evaluated, to reduce bias in their behavior. Question 3 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters 𝜇 = 71 μ=71 and 𝜎 2 = 6.25 σ 2 =6.25. a. What percentage of 25-year-old men are over 6 feet 2 inches tall? b. What percentage of men in the 6-footer club are over 6 feet 5 inches? Answer 1 a. To calculate the percentage of men, we first calculate the probability: 𝑃 ( Height of a 25-year-old man is over 6 feet 2 inches ) = 𝑃 ( 𝑋 > 74 ) P(Height of a 25-year-old man is over 6 feet 2 inches)=P(X>74) 𝑃 ( 𝑋 > 74 ) = 𝑃 ( 𝑍 > 𝑋 − 𝜇 𝜎 ) = 𝑃 ( 𝑍 > 74 − 71 2.5 ) = 𝑃 ( 𝑍 > 1.2 ) . P(X>74)=P(Z> σ X−μ ​ )=P(Z> 2.5 74−71 ​ )=P(Z>1.2). Using the standard normal table: 𝑃 ( 𝑍 > 1.2 ) = 1 − Φ ( 1.2 ) = 1 − 0.8849 = 0.1151. P(Z>1.2)=1−Φ(1.2)=1−0.8849=0.1151. Thus, 11.51% of 25-year-old men are over 6 feet 2 inches tall. b. 𝑃 ( Height of 25-year-old man is above 6 feet 5 inches given that he is above 6 feet ) = 𝑃 ( 𝑋 > 77 ) 𝑃 ( 𝑋 > 72 ) . P(Height of 25-year-old man is above 6 feet 5 inches given that he is above 6 feet)= P(X>72) P(X>77) ​ . 𝑃 ( 𝑋 > 77 ) = 𝑃 ( 𝑍 > 77 − 71 2.5 ) = 𝑃 ( 𝑍 > 2.4 ) . P(X>77)=P(Z> 2.5 77−71 ​ )=P(Z>2.4). Using the standard normal table: 𝑃 ( 𝑍 > 2.4 ) = 1 − Φ ( 2.4 ) = 1 − 0.9918 = 0.0082. P(Z>2.4)=1−Φ(2.4)=1−0.9918=0.0082. 𝑃 ( 𝑋 > 72 ) = 𝑃 ( 𝑍 > 72 − 71 2.5 ) = 𝑃 ( 𝑍 > 0.4 ) . P(X>72)=P(Z> 2.5 72−71 ​ )=P(Z>0.4). 𝑃 ( 𝑍 > 0.4 ) = 1 − Φ ( 0.4 ) = 1 − 0.6554 = 0.3446. P(Z>0.4)=1−Φ(0.4)=1−0.6554=0.3446. 𝑃 ( 𝑋 > 77 ∣ 𝑋 > 72 ) = 0.0082 0.3446 = 0.024. P(X>77∣X>72)= 0.3446 0.0082 ​ =0.024. Thus, 2.4% of men in the 6-footer club are over 6 feet 5 inches. Question 4 Suppose that 𝑋 X is a normal random variable with mean 5. If 𝑃 ( 𝑋 > 9 ) = 0.2 P(X>9)=0.2, approximately what is Var ( 𝑋 ) Var(X)? Answer 1 𝑋 X is a normal random variable with: 𝜇 = 5 and 𝑃 ( 𝑋 > 9 ) = 0.2. μ=5andP(X>9)=0.2. Let 𝑋 X's variance be 𝜎 2 σ 2 . Then: 𝑃 ( 𝑋 > 9 ) = 0.2    ⟹    𝑃 ( 𝑍 > 9 − 5 𝜎 ) = 0.2. P(X>9)=0.2⟹P(Z> σ 9−5 ​ )=0.2. 𝑃 ( 𝑍 > 4 𝜎 ) = 0.2    ⟹    Φ ( 4 𝜎 ) = 0.8. P(Z> σ 4 ​ )=0.2⟹Φ( σ 4 ​ )=0.8. From the standard normal table: Φ − 1 ( 0.8 ) = 0.845    ⟹    4 𝜎 = 0.845    ⟹    𝜎 = 4 0.845 ≈ 4.73. Φ −1 (0.8)=0.845⟹ σ 4 ​ =0.845⟹σ= 0.845 4 ​ ≈4.73. Thus: Var ( 𝑋 ) = 𝜎 2 = ( 4.73 ) 2 ≈ 22.37. Var(X)=σ 2 =(4.73) 2 ≈22.37. Question 5 Given a standard normal distribution, find the area under the curve that lies: (a) To the left of 𝑧 = − 1.39 z=−1.39. (b) To the right of 𝑧 = 1.96 z=1.96. (c) Between 𝑧 = − 2.16 z=−2.16 and 𝑧 = − 0.65 z=−0.65. (d) To the left of 𝑧 = 1.43 z=1.43. (e) To the right of 𝑧 = − 0.89 z=−0.89. (f) Between 𝑧 = − 0.48 z=−0.48 and 𝑧 = 1.74 z=1.74. Answer 1 (a) 𝑃 ( 𝑍 < − 1.39 ) = 0.0823. P(Z<−1.39)=0.0823. (b) 𝑃 ( 𝑍 > 1.96 ) = 1 − Φ ( 1.96 ) = 1 − 0.9750 = 0.025. P(Z>1.96)=1−Φ(1.96)=1−0.9750=0.025. (c) 𝑃 ( − 2.16 < 𝑍 < − 0.65 ) = Φ ( − 0.65 ) − Φ ( − 2.16 ) = 0.2578 − 0.0154 = 0.2424. P(−2.16 − 0.89 ) = 1 − Φ ( − 0.89 ) = 1 − 0.1867 = 0.8133. P(Z>−0.89)=1−Φ(−0.89)=1−0.1867=0.8133. (f) 𝑃 ( − 0.48 < 𝑍 < 1.74 ) = Φ ( 1.74 ) − Φ ( − 0.48 ) = 0.9599 − 0.3156 = 0.6435. P(−0.48