I met this unforgettable character when I was just minutes old. This person means a great deal to me. He is still in my life‚ but not as I always remember. My father and I did everything together. I was most definitely daddy’s little girl. My father had all the typical dad features. He was big and strong and made our home feel safe. I can remember cuddling in his lap watching television. His arms were so thick and mighty. He made me feel very secure. I remember helping him plant the garden. My
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Mary Pittman Mr. Zemp English 101 January 20‚ 2010 Unforgettable Night What started out as a normal night‚ ended up as a night from hell. It was the day of graduation and everything was perfect. This day marked the beginning of the summer before my senior year. I had already made plans to go on trips to the beach with my family and friends. My friend‚ Bailes‚ was in town for graduation‚ as a surprise. She moved to Ohio her sophomore year and came to South Carolina when she could‚ which was
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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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An unforgettable memory The silence was so dense and heavy I could almost feel it but it was always like that between me and dad. He had told me to go grocery shopping with him because we had come back after the summer vacation and there was no proper and edible grocery at home‚ so we were headed to the local store. Dad was really sick and weak after the flight back from our homeland so I made myself do all the shopping quickly so that he wouldn’t have to tire himself. And whenever I am in
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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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