University:
Shippensburg University of Pennsylvania
Course:
MAT 211 | Calculus I
Academic year:

2024
Views:

231

Pages:

4
Author:

Herbosambx

Question 1 Random variables Given independent random variables with means and standard deviations as shown, find the mean and standard deviation of: 𝑋 + 𝑌 X+Y. Answer 1 The mean and standard deviation are as follows: 𝐸 ( 𝑋 + 𝑌 ) = 𝐸 ( 𝑋 ) + 𝐸 ( 𝑌 ) E(X+Y)=E(X)+E(Y) 𝐸 ( 𝑋 + 𝑌 ) = 80 + 12 = 92 E(X+Y)=80+12=92 𝑉 ( 𝑋 + 𝑌 ) = 𝑉 ( 𝑋 ) + 𝑉 ( 𝑌 ) V(X+Y)=V(X)+V(Y) 𝑆 𝐷 ( 𝑋 + 𝑌 ) = 𝑉 ( 𝑋 ) + 𝑉 ( 𝑌 ) = 1 2 2 + 3 2 = 144 + 9 = 153 ≈ 12.3693 SD(X+Y)= V(X)+V(Y) = 12 2 +3 2 = 144+9 = 153 ≈12.3693 Question 2 If 𝑋 X is a normal random variable with parameters 𝜇 = 10 and 𝜎 2 = 36 , μ=10andσ 2 =36, compute 𝑃 ( 𝑋 < 20 ) P(X<20). Answer 1 Given that 𝑋 X is a mean-containing normal random variable 𝜇 = 10 μ=10 and variance 𝜎 2 = 36 σ 2 =36: The following probability is calculated using the standard normal distribution: 𝑍 = 𝑋 − 𝜇 𝜎 ⟹ 𝑍 = 20 − 10 6 = 10 6 = 5 3 Z= σ X−μ ⟹Z= 6 20−10 = 6 10 = 3 5 Using the normal table: 𝑃 ( 𝑋 < 20 ) = 𝑃 ( 𝑍 < 5 3 ) = Φ ( 5 3 ) = 0.9522 P(X<20)=P(Z< 3 5 )=Φ( 3 5 )=0.9522 Question 3 Can a sample have a standard deviation of zero? Answer 1 Explanation: It is feasible, but only if a sample is made up of identical data. Example dataset: 5 , 5 , 5 , 5 , 5 , 5 , 5 , 5 , 5 , 5 5,5,5,5,5,5,5,5,5,5. The sample standard deviation will be 0 if all sample values are identical. Question 4 For a population with a mean of 𝜇 = 100 μ=100 and a standard deviation of 𝜎 = 20 σ=20, find the 𝑋 X-values: 𝑧 = + 0.75 z=+0.75 Answer 1 If the 𝑧 z-score is 0.75, find the value of the random variable 𝑋 X: 𝑧 = 𝑋 − 𝜇 𝜎 ⟹ 𝑋 = 𝜇 + 𝑧 ⋅ 𝜎 z= σ X−μ ⟹X=μ+z⋅σ 𝑋 = 100 + ( 0.75 ) ( 20 ) = 100 + 15 = 115 X=100+(0.75)(20)=100+15=115 The random variable 𝑋 X will have the value 115 115. Question 5 How many different 10-letter words (real or imaginary) can be formed from the following letters: 𝐻 , 𝑇 , 𝐺 , 𝐵 , 𝑋 , 𝑋 , 𝑇 , 𝐼 , 𝑁 , 𝐽 H,T,G,B,X,X,T,I,N,J Answer 1 Approach: The number of different 10-letter words is determined by accounting for repeated letters. The letters are 𝐻 , 𝑇 , 𝐺 , 𝐵 , 𝑋 , 𝑋 , 𝑇 , 𝐼 , 𝑁 , 𝐽 H,T,G,B,X,X,T,I,N,J. Total letters = 10 Repetitions: 𝑋 X appears twice, 𝑇 T appears twice. The formula for permutations accounting for repetitions is: 𝑛 ! 𝑛 1 ! ⋅ 𝑛 2 ! n 1 !⋅n 2 ! n! where 𝑛 = 10 n=10, 𝑛 1 = 2 n 1 =2 (for 𝑋 X), 𝑛 2 = 2 n 2 =2 (for 𝑇 T). Total permutations = 10 ! 2 ! ⋅ 2 ! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ( 2 ⋅ 1 ) ( 2 ⋅ 1 ) Total permutations= 2!⋅2! 10! = (2⋅1)(2⋅1) 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 = 3628800 4 = 907200 = 4 3628800 =907200 Thus, the total number of 10-letter words = 907,200.