Final Exam Review Questions Solutions Guide You will probably want to PRINT THIS so you can carefully check your answers. Be sure to ask your instructor if you have questions about any of the solutions given below. 1. Explain the difference between a population and a sample. In which of these is it important to distinguish between the two in order to use the correct formula? mean; median; mode; range; quartiles; variance; standard deviation. Solution: A sample is a subset of a population. A population
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Microorganisms as Pathogens To be considered a pathogen it must: Gain entry Colonise the tissues Resist the defences Cause damage to the tissues Pathogens include bacteria‚ viruses and fungi How do microorganisms enter the body Many pathogens enter through the gas exchange system (including ones that cause flu and TB) Food and water can carry pathogens into the stomach and intestines via the mouth and into the digestive system (including ones that cause cholera)
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What activities are defined as a crime? Activities that are defined as a crime are activities or behaviors that a society deems as morally and inherently wrong. My opinion‚ and the opinion of the majority of society‚ is an activity being labelled as morally wrong in the circumstance of crime can be an act that does‚ or has the possibility‚ of emotionally‚ mentally‚ or physically harming others. Activities that are defined as a crime depend on the norms and beliefs of a society‚ so an activity being
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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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decimal places) 2. Find the value of z if the area under a Standard Normal curve a) to the right of z is 0.3632; b) to the left of z is 0.1131; c) between 0 and z‚ with z > 0‚ is 0.4838; d) between -z and z‚ with z > 0‚ is 0.9500. Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table) b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table) c ) the area between 0 to z is 0.4838‚ z = 2.14 d) the area to the
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There are a number of problems that may arise when supporting learning activities which could relate to any one of the following: the learners‚ the learning activities‚ the learning resources or the learning environment. In terms of the learners you may find that they are unable to achieve the learning objectives for a variety of reasons. You may experience bad behaviour amongst the group‚ from one or more of the children. If bad behaviour occurs the adult must intervene immediately to stop it disrupting
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random will be between 19 and 31 is about 0.95. This area (probability) is shown fir the X values and for the z values. σ = 3 0.95 σ = 1 0.95 X 19 25 31 -2 0 +2 Normal curve showing Standard normal curve showing area between 19 and 31 area between -2 and +2 Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100
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Application of Normal mode calculations in optical spectroscopy | Phym 221 – Assignment 4 1. Introduction Normal modes are used to describe the different vibration motion in molecules. There are different types of modes for molecules in different motions and each has a certain symmetry associated with it. 2. Overview of Normal Modes Generally‚ normal modes are independent atoms in a molecule that are in motion such that they do not disturb the motion of the other molecules. Normal modes as implied
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Normal Distribution Normal distribution is a statistics‚ which have been widely applied of all mathematical concepts‚ among large number of statisticians. Abraham de Moivre‚ an 18th century statistician and consultant to gamblers‚ noticed that as the number of events (N) increased‚ the distribution approached‚ forming a very smooth curve. He insisted that a new discovery of a mathematical expression for this curve could lead to an easier way to find solutions to
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Table Z: Areas under the standard normal curve (negative Z) Second decimal place in z 0.06 0.05 0.04 0.03 0.09 0.08 0.07 0.02 0.01 0.0001 0.0001 0.0002 0.0002 0.00 * 0.0000 0.0001 0.0001 0.0002 0.0002 z -3.9 -3.8 -3.7 -3.6 -3.5 0.0001 0.0001 0.0001 0.0002 0.0001 0.0001 0.0001 0.0002 0.0001 0.0001 0.0001 0.0002 0.0001 0.0001 0.0001 0.0002 0.0001 0.0001 0.0001 0.0002 0.0001 0.0001 0.0001 0.0002 0.0001 0.0001 0.0001 0
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