variance is 846‚ what is the standard deviation? Solution: standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data 34‚ 38‚ 22‚ 21‚ 29‚ 37‚ 40‚ 41‚ 22‚ 20‚ 49‚ 47‚ 20‚ 31‚ 34‚ 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6 This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot? 45 40 35 30
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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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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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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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Chapter 13: Chi-Square Applications SHORT ANSWER 1. When samples of size n are drawn from a normal population‚ the chi-square distribution is the sampling distribution of = ____________________‚ where s2 and are the sample and population variances‚ respectively. ANS: PTS: 1 OBJ: Section 13.2 2. Find the chi-square value for each of the right-tail areas below‚ given that the degrees of freedom are 7: A) 0.95 ____________________ B) 0.01 ____________________ C) 0.025 ____________________
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probably not attributable to chance is: (Points : 1) | Type I error Type II error Statistical significance In the semi-quartile range | 5. A score that is likely to fall into the middle 68% of scores of a normal distribution will fall inside these values: (Points : 1) | . +/- 3 standard deviations +/- 2 standard deviations +/- 1 standard deviation semi-quartile range | 6. It is important to assess the magnitude or strength
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Describing Data [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] FREE RESPONSE Use the given data set of test grades from a college statistics class for this question. 85 72 64 65 98 78 75 76 82 80 61 92 72 58 65 74 92 85 74 76 77 77 62 68 68 54 62 76 73 85 88 91 99 82 80 74 76 77 70 60 A. Construct two different graphs of these data B. Calculate the five-number summary and the mean and standard deviation of the data. C. Describe the distribution of the data‚ citing both
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Business Statistics MGSC-372 Review Normal Distribution The Normal Distribution aka The Gaussian Distribution The Normal Distribution y 1 f ( x) e 2 1 x 2 2 x Areas under the Normal Distribution curve -3 -2 - 68% 95% 99.7% + +2 +3 X = N( ‚ 2 ) Determining Normal Probabilities Since each pair of values for and represents a different distribution‚ there are an infinite number of possible normal distributions. The number of statistical tables
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Marks: 1 Assume that X has a normal distribution‚ and find the indicated probability. The mean is μ = 60.0 and the standard deviation is σ = 4.0. Find the probability that X is less than 53.0. Choose one answer. a. 0.5589 b. 0.0401 c. 0.9599 d. 0.0802 Question2 Marks: 1 Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normal distribution. Weights of eggs: 95% confidence;
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