Student Exploration: Sight vs. Sound Reactions Vocabulary: histogram‚ mean‚ normal distribution‚ range‚ standard deviation‚ stimulus Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Most professional baseball pitchers can throw a fastball over 145 km/h (90 mph). This gives the batter less than half a second to read the pitch‚ decide whether to swing‚ and then try to hit the ball. No wonder hitting a baseball is considered one of the hardest things to do in sports! 1. What
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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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Pages C H A P T E R 6 The Normal Distribution Objectives Outline After completing this chapter‚ you should be able to 1 2 3 Identify distributions as symmetric or skewed. 4 Find probabilities for a normally distributed variable by transforming it into a standard normal variable. Introduction 6–1 Normal Distributions Identify the properties of a normal distribution. Find the area under the standard normal distribution‚ given various z values. 5 Find
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NORMAL DISTRIBUTION 1. Find the distribution: a. b. c. d. e. f. following probabilities‚ the random variable Z has standard normal P (0< Z < 1.43) P (0.11 < Z < 1.98) P (-0.39 < Z < 1.22) P (Z < 0.92) P (Z > -1.78) P (Z < -2.08) 2. Determine the areas under the standard normal curve between –z and +z: ♦ z = 0.5 ♦ z = 2.0 Find the two values of z in standard normal distribution so that: P(-z < Z < +z) = 0.84 3. At a university‚ the average height of 500 students of a course is 1.70 m; the standard
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comorbidities)‚ and depression. Random samples of 20 healthy individuals were selected from three geographic locations: Florida‚ New York‚ and North Carolina. Then‚ each was given a standardized test to measure depression (higher test scores indicate higher levels of depression). Similarly‚ random samples of 20 individuals with one or more comorbidities (arthritis‚ hypertension‚ and/or heart ailment) were taken from the three geographic locations. They were also given the standardized test to measure
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Normal Distribution It is important because of Central Limit Theorem (CTL)‚ the CTL said that Sum up a lot of i.i.d random variables the shape of the distribution will looks like Normal. Normal P.D.F Now we want to find c This integral has been proved that it cannot have close form solution. However‚ someone gives an idea that looks stupid but actually very brilliant by multiply two of them. reminds the function of circle which we can replace them to polar coordinate Thus Mean
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( 4 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
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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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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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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(0.4772) = 0.9554 Therefore the probability that a measurement selected at 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
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