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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deviation of any normal random variable is always equal to one. Answer: False Type: Concept Difficulty: Easy 2. For any normal random variable‚ the probability that the random variable will equal one is always zero. Answer: True Type: Concept Difficulty: Medium 3. The graph of a standard normal random variable is always symmetric. Answer: True Type: Concept Difficulty: Easy 4. The formula will convert any normal distribution into the “standard normal distribution
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1. Cu = 24-11 = $13 Co = 11-7 = $4 Critical ratio = 13/(13+4) = 0.7647 μ = 30‚000 σ = 10‚000 Using normal distribution function (=norminv(0.7647‚30000‚10000))‚ the optimum order quantity is 37‚216 jerseys to maximize profit. 2. Quantity = 32‚000 First‚ we normalize the order quantity to find the z-statistic z=Q-μσ=32‚000-25‚00010‚000=0.7 We then look up the standard normal loss function. The expected lost sale is given by. Lz=0.1429 Therefore‚ the expected lost sales = 10‚000 *
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D‚ E) with the following functional Dependencies: {BC! ADE‚ D! B}. a. Find all candidate keys. b. Identify the best normal form that R satisfies (1NF‚ 2NF‚ 3NF‚ or BCNF). c. If the relation is not in BCNF‚ decompose it until it becomes BCNF. At each step‚ identify a new relation‚ decompose and re-compute the keys and the normal forms they satisfy. 1 . Answer any five of the questions below. What do your understand by normalization. And also explain
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Fruit Fly Lab Alycia Fletcher Biology IB HL March 25th 2010 Fruit Fly Lab Introduction Genes can either be sex-linked or autosomal. If a gene appears mostly in one sex chances are the gene is sex-linked and if it appears frequently in both sexes it is most likely autosomal. Using Drosophila melanogaster‚ also known as the fruit fly‚ we will determine whether the gene is sex-linked or autosomal. Drosophila melanogasters have a relatively short life span and are an excellent organism
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and 6? (A) 0.550 (B) 0.575 (C) 0.600 (D) 0.625 (E) 0.650 2. Which of the following statements about a normal distribution is true? (A) The value of µ must always be positive. (B) The value of σ must always be positive. (C) The shape of a normal distribution depends on the value of µ. (D) The possible values of a standard normal variable range from −3.49 to 3.49. (E) The area under a normal curve depends on the value of σ. 3. A variable X follows a uniform distribution‚ as shown below: The
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twelve (12) everyone thought it was just another ‘boys will be boys’ and accepted that pulling pranks at that age was normal. Normal‚ until I had done the unthinkable. I did what no one had expected me to do: using fake animals to frighten people into thinking it was real was normal‚ switching my sister’s shampoo with dishwashing liquid was normal‚ giving my teacher a fake apple was normal; everyone expected and accepted my behavior but I was not done. It took everyone by surprise. It was my thirteenth
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radius decreased‚ characteristic of an obstructive pulmonary problem Activity 2 Comparative Spriometry Chart 2: Spirometery Results Patient Type TV (ml) ERV (ml) IRV (ml) RV (ml) FVC (ml) TLC (ml) FEV1 (ml) FEV1 (%) Normal 500 1500 3000 1000 5000 6000 4000 80% Emphysema 500 750 2000 2750 3250 6000 1625 50% Acute asthma attack 300 750 2700 2250 3750 6000 1500 40% Plus inhaler 500 1500 2800 1200 4800
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A Derivation of an Upper Bound for the Number of Configurations of an n×n×n×n Rubik ’s Cube By David Smith 1. Introduction C4(n) is a formula for an upper bound of the number of distinguishable configurations of an n×n×n×n Rubik ’s Cube‚ which will be derived in this paper. It will be assumed that the reader is familiar with a 4-dimensional Rubik ’s Cube. Online‚ one can find the free computer program Magic Cube 4D‚ developed by Melinda Green‚ Don Hatch‚ and Jay Berkenbilt‚ which is a completely
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adipose tissue is forming (normal during teen development)‚ normal pinkish gray coloring‚ soft tissue. Thyroid 30gm Right Lung Left Lung Heart Gall Bladder Liver Spleen Stomach Small Intestine 490gm unremarkable pleural surfaces are translucent‚ smooth and glistening 450gm the parenchyma is pink and normal 275mg Normal weight surfaces are glistening and unremarkable. pericardium intact 2 ounces pear shaped hollow organ‚ normal no gallstones
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