University:
Florida Gateway College
Course:
MAC 2312 | Calculus II
Academic year:

2024
Views:

413

Pages:

2
Author:

Ruby Bates

Question 1 a) If the standard deviation is 0, then the variance is b) The Empirical Rule states that approximately what percentage of the data lies within one standard deviation of the mean? c) Chebyshev's theorem states that there is at least what percentage of the data lies within two standard deviations of the mean? Answer: Step 1: a) Variance = (Standard Deviation) 2 = 0 2 = 0 2 =0 2 =0. b) The Empirical Rule states that approximately 68% of the data lies within one standard deviation of the mean. Step 2: c) Chebyshev's theorem states that at least 1 − 1 𝑘 2 1− k 2 1 of the data lies within 𝑘 k standard deviations of the mean. For 𝑘 = 2 k=2: 1 − 1 2 2 = 1 − 1 4 = 3 4 = 0.75 = 75 % . 1− 2 2 1 =1− 4 1 = 4 3 =0.75=75%. Answer: a) 0 b) 68 c) 75 Question 2 Find the mean, median, and mode of the given set of raw data. 1.2, 0.7, 1.5, 1.0, 0.7, 1.9, 1.7, 1.2 0.8, 1.7, 1.3, 2.3, 0.9, 2.0, 1.7, 1.5, 2.2 1.45 Answer: Step 1: Given data: 𝑥 = { 1.2 , 0.7 , 1.5 , 1.0 , 0.7 , 1.9 , 1.7 , 1.2 , 0.8 , 1.7 , 1.3 , 2.3 , 0.9 , 2.0 , 1.7 , 1.5 , 2.2 } x={1.2,0.7,1.5,1.0,0.7,1.9,1.7,1.2,0.8,1.7,1.3,2.3,0.9,2.0,1.7,1.5,2.2}. Σ 𝑥 = 27.55 , 𝑛 = 19. Σx=27.55,n=19. Step 2: Mean Mean = Σ 𝑥 𝑛 = 27.55 19 = 1.45. Mean= n Σx = 19 27.55 =1.45. Median Sort data in ascending order: 0.7, 0.7, 0.8, 0.9, 1.0, 1.2, 1.2, 1.3, 1.5, 1.5, 1.7, 1.7, 1.7, 1.7, 1.9, 2.0, 2.2, 2.3. The median is the ( 𝑛 + 1 2 ) 𝑡 ℎ ( 2 n+1 ) th observation. Median = 1.5. Median=1.5. Mode The mode is the value that appears most frequently. Mode = { 1.7 , 1.5 , 0.7 } {1.7,1.5,0.7}. Final Answer: Mean: 1.45 Median: 1.5 Mode: 1.7, 1.5, 0.7 Question 3 Supplier delivery times follow a normal distribution with a mean 𝜇 = 48.2 μ=48.2 minutes and a standard deviation 𝜎 = 13.4 σ=13.4 minutes. What is the probability that the mean time of 5 deliveries exceeds 60 minutes? Answer: Step 1: Let 𝑋 X be the delivery times that follows normal distribution 𝑁 ( 𝜇 = 48.2 , 𝜎 = 13.4 ) N(μ=48.2,σ=13.4). Probability that mean time of 5 deliveries exceeds 60 minutes: 𝑃 ( 𝑋 ˉ ≥ 60 ) = 𝑃 ( 𝑍 ≥ 𝑋 ˉ − 𝜇 𝜎 / 𝑛 ) P( X ˉ ≥60)=P(Z≥ σ/ n X ˉ −μ ) Step 2: Substitute: 𝜎 𝑛 = 13.4 5 = 6. n σ = 5 13.4 =6. 𝑃 ( 𝑍 ≥ 60 − 48.2 6 ) = 𝑃 ( 𝑍 ≥ 1.97 ) . P(Z≥ 6 60−48.2 )=P(Z≥1.97). From standard normal table: 𝑃 ( 𝑍 ≥ 1.97 ) = 1 − 𝑃 ( 𝑍 ≤ 1.97 ) = 1 − 0.9756 = 0.0244. P(Z≥1.97)=1−P(Z≤1.97)=1−0.9756=0.0244. Final Answer: The probability that the mean time of 5 deliveries exceeds 60 minutes is 0.0244 (2.44%).