University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2024
Views:

301

Pages:

3
Author:

Beckett Gray

Question 1 To indicate: The area under the appropriate normal curve would be determined to approximate the binomial probability 𝑃 ( 7 ≤ 𝑋 ≤ 10 ) P(7≤X≤10). Answer: Using continuity correction, the binomial probability 𝑃 ( 7 ≤ 𝑋 ≤ 10 ) P(7≤X≤10) is approximated as: 𝑃 ( 6.5 ≤ 𝑋 ≤ 10.5 ) P(6.5≤X≤10.5) This involves adjusting the discrete binomial values by 0.5 0.5 to approximate a continuous normal distribution. Question 2 The quality-control inspector checks defective syringes: 𝑝 = 0.01 p=0.01, 𝑛 = 8 n=8 What is the expected number of defective syringes? 𝜇 = 𝑛 𝑝 = 8 × 0.01 = 0.08 syringes. μ=np=8×0.01=0.08syringes. Probability the batch will be accepted ( 𝑃 ( 𝑋 ≤ 1 ) P(X≤1)): Using the binomial probability formula: 𝑃 ( 𝑋 ≤ 1 ) = 𝑃 ( 𝑋 = 0 ) + 𝑃 ( 𝑋 = 1 ) P(X≤1)=P(X=0)+P(X=1) 𝑃 ( 𝑋 = 0 ) = ( 8 0 ) ( 0.01 ) 0 ( 0.99 ) 8 = 0.9227 P(X=0)=( 0 8 )(0.01) 0 (0.99) 8 =0.9227 𝑃 ( 𝑋 = 1 ) = ( 8 1 ) ( 0.01 ) 1 ( 0.99 ) 7 = 8 ( 0.01 ) ( 0.9321 ) = 0.0746 P(X=1)=( 1 8 )(0.01) 1 (0.99) 7 =8(0.01)(0.9321)=0.0746 𝑃 ( 𝑋 ≤ 1 ) = 0.9227 + 0.0746 = 0.9973 P(X≤1)=0.9227+0.0746=0.9973 Thus, the probability that the batch will be accepted is 0.9973 0.9973. Question 3 Assume a Poisson distribution with 𝜆 = 5.0 λ=5.0. Find 𝑃 ( 𝑋 = 1 ) P(X=1). Answer: Using the Poisson probability formula: 𝑃 ( 𝑋 = 𝑥 ) = 𝑒 − 𝜆 𝜆 𝑥 𝑥 ! P(X=x)= x! e −λ λ x For 𝑥 = 1 x=1 and 𝜆 = 5 λ=5: 𝑃 ( 𝑋 = 1 ) = 𝑒 − 5 5 1 1 ! = 0.0067 ⋅ 5 1 = 0.0337 P(X=1)= 1! e −5 5 1 = 1 0.0067⋅5 =0.0337 Question 4 Numerical relationship between 𝑝 p and 𝑞 q in a binomial distribution: Answer: The relationship is: 𝑝 + 𝑞 = 1 or 𝑞 = 1 − 𝑝 . p+q=1orq=1−p. Here, 𝑝 p is the probability of success, and 𝑞 q is the probability of failure. Question 5 How many international students are expected in a cluster of 40 students ( 𝑛 = 40 , 𝑝 = 0.06 n=40,p=0.06)? Answer: The mean 𝜇 μ is given by: 𝜇 = 𝑛 𝑝 = 40 × 0.06 = 2.4 students. μ=np=40×0.06=2.4students. The standard deviation 𝜎 σ is: 𝜎 = 𝑛 𝑝 𝑞 = 40 × 0.06 × ( 1 − 0.06 ) σ= npq = 40×0.06×(1−0.06) 𝜎 = 40 × 0.06 × 0.94 = 2.256 ≈ 1.5020 σ= 40×0.06×0.94 = 2.256 ≈1.5020 Final Answer: Expected: 2.4 students Standard deviation: 1.5020 students