and what he saw a dead body on the floor and that was his wife. It was witching hour on a dark moonless night. He was holding his wife body in his arms‚ kissing and crying. Suddenly‚ he heard their voices downstairs again. He stopped crying for a moment. He didn’t know that was his last day of his life. The phone rang once again and answered itself saying” remembered Oct 4th at 2 am”. John knew that day and time very well because he accidentally murdered a married couple on this day. He and his wife
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baked cookies filled the air‚ as the jolly music played in the background. People are wearing colourfully knitted scarfs and sweaters. The golden sun was always shining in the clear blue sky. Those were the days I will never forget. That special moment when nana pulled out the golden‚ crispy turkey is when you know it’s Christmas time. The lit-up streets were full of dancing‚ bright Santa’s. Every house had flickering‚ blinding lights that made every minute the most exciting yet. There is always
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Normal Distribution:- A continuous random variable X is a normal distribution with the parameters mean and variance then the probability function can be written as f(x) = - < x < ‚ - < μ < ‚ σ > 0. When σ2 = 1‚ μ = 0 is called as standard normal. Normal distribution problems and solutions – Formulas: X < μ = 0.5 – Z X > μ = 0.5 + Z X = μ = 0.5 where‚ μ = mean σ = standard deviation X = normal random variable Normal Distribution Problems and Solutions – Example
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Mean of a log normal random variable: Theorem 1: Suppose Y = ln X is a normal distribution with mean m and variance v‚ then X has mean exp( m + v /2 ) Proof: The density function of Y= ln X Therefore the density function of X is given by Using the change of variable x = exp(y)‚ dx = exp(y) dy‚ We have = Note that the integral inside is just the density function of a normal random variable with mean (m-v) and variance v. By definition‚ the integral evaluates to be 1. Proof of Black Scholes
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ROP based on Normal Distribution of LT demand -Example No. 1 Example : Suppose that the manager of a construction supply house determined from historical records that demand for sand during leadtime averages 50 tons. In addition‚ suppose the manager determined that the demand during leadtime could be described by a normal distribution that has a mean of 50 tons and a standard deviation of 5 tons. Answer the following questions‚ assuming that the manager is willing to accept a stockout risk of no
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Problem Sheet - I 1. Researcher conducted by a tobacco company indicates that the relative frequency distribution of tar content of its newly developed low-tar cigarette has a mean equal to 3.9 milligrams of tar per cigarette and a standard deviation equal to 1.0 milligram. Suppose a sample of 100 low-tar cigarettes is randomly selected from a day’s production and the tar content is measured in each. Assuming that the tobacco company’s claim is true‚ what is the probability that the mean
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13. Variance and Standard Deviation (expected). Using the data from problem 13‚ calculate the variance and standard deviation of the three investments‚ stock‚ corporate bond‚ and government bond. If the estimates for both the probabilities of the economy and the returns in each state of the economy are correct‚ which investment would you choose considering both risk and return? Why? ANSWER Variance of Stock = 0.10 x (0.25 – 0.033)2 + 0.15 x (0.12 – 0.033)2 + 0.50 x (0.04 – 0.033)2 + 0
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A population of measurements is approximately normally distributed with mean of 25 and a variance of 9. Find the probability that a measurement selected at random will be between 19 and 31. Solution: The values 19 and 31 must be transformed into the corresponding z values and then the area between the two z values found. Using the transformation formula from X to z (where µ = 25 and σ √9 = 3)‚ we have z19 = (19 – 25) / 3 = -2 and z31 = (31 - 25) / 3 = +2 From the area between z =±2 is 2(0
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STAT11-111 Business Statistics WEEK 3 and WEEK 4: WORKSHOPS ______________________________________________ This workshop is to be completed during Week 3 and Week 4 workshops and will depend on how quickly we get through the lecture material. Part A: Week 3 Exercise 4.2‚ Exercise 4.4. Exercise 4.5‚ Exercise 4.6‚ Exercise 4.7. Your tutor will discuss these 3 questions with you in the class. Exercise 4.8‚ Exercise 4.9. Exercise 4.10‚ Exercise 4.12‚ Exercise 4.14. Exercise 4.15‚ Exercise
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Assignment: Interpreting Correlational Findings Following are brief summaries of correlational findings‚ in which variables were found to be significantly associated with each other. Your task is to determine which of the three major causal models (i.e.‚ interpretations) could account for each finding. Indicate in the table below‚ by placing an X in the appropriate space‚ which of these three models could provide a possible explanation. Place an X in the space only if you judge the causal
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