Examination Name: Course: Date: Final Examination in Statistics (M.A.Ed./M.A.N.) 1.The scores of 15 masteral students in Statistics were 80‚85‚78‚90‚91‚98‚95‚98‚95‚74‚71‚72‚98‚99‚and 87. Find the measures of central tendency‚ the range‚ the variance‚ and the standard deviation. 2. In the performance evaluation of teachers‚ if the dean’s evaluation is given a weight of 5‚ self-evaluation is 2‚ peer’s evaluation is 2‚ and student’s evaluation is 1 and the teacher’s rating is 90‚ 95
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Use Definition 4.3 on page 120 to find the variance of the random variable X of Exercise 4.7 on page 117. 4.7 By investing in a particular stock‚ a person can make a profit in one year of $4000 with probability 0.3 or take a loss of $1000 with probability 0.7. What is this person’s expected gain? 4.37 A dealer’s profit‚ in units of $5000‚ on a new automobile is a random variable X having the density function given in Exercise 4.12 on page 117. Find the variance of X. 4.12 If a dealer’s profit‚ in units
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of hypothesis for a single mean (σ unknown): t= 1 ¯ −µ X √ S/ n Z-test of hypothesis proportion: for a single Z-test for the difference between two means (variances known): p−π Z∼ = Z= π(1−π) n t-test for the difference between two means (variances unknown): t= 1 n1 + + (¯ x1 − x ¯2 ) ± tn1 +n2 −2 1 n2 Pooled variance estimator: Sp2 = σ12 n1 t= sd x ¯d ± tn−1 √ n ¯ d − µd X √ Sd / n Z= (P1 − P2 ) − (π1 − π2 ) P (1 − P ) R1 + R 2 n1 + n2 (p1 − p2 ) ± z χ2 test of association:
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Male 382 3.01 2.648 .135 Female 524 3.00 2.497 .109 Independent Samples Test Levene’s Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper Hours per day watching TV Equal variances assumed .221 .638 .068 904 .946 .012 .172 -.327 .350 Equal variances not assumed .067 792.603 .947 .012 .174 -.330 .353 There is no significant difference between male(mean=3.01
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Gaussian Mixture Models∗ Douglas Reynolds MIT Lincoln Laboratory‚ 244 Wood St.‚ Lexington‚ MA 02140‚ USA dar@ll.mit.edu Synonyms GMM; Mixture model; Gaussian mixture density Definition A Gaussian Mixture Model (GMM) is a parametric probability density function represented as a weighted sum of Gaussian component densities. GMMs are commonly used as a parametric model of the probability distribution of continuous measurements or features in a biometric system‚ such as vocal-tract related
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that would be good investments for Beta Management. Her analyst has provided her with two different buckets of stocks for her to decide on. With her inexperience in the financial industry she is unsure as to which bucket to decide to invest in. Analysis: It is very important in this case to look into what the goal of the fund is. In this scenario‚ Ms. Wolfe’s goal is to make more than she could in a bank‚ but also keep her risk exposure as low as possible. This is what we will be making the decision
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Assignment 12 February 2014 1. Analysis a Business Issue. Introduction New Look Jacket Inc. (NLJ) specializes in the production of Nylon Jackets and Leather Jackets. The company delivers successful financial records at the end of the 2012 fiscal year with the net income of $ 417‚100‚ which is $170‚850 greater than the net income budgeted for the 2012 fiscal year despite that the company operations goes through some turmoil. A more detail variance shows that the external factor largely
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Wiener Process Ito ’s Lemma Derivation of Black-Scholes Solving Black-Scholes Introduction to Financial Derivatives Understanding the Stock Pricing Model 22M:303:002 Understanding the Stock Pricing Model 22M:303:002 Wiener Process Ito ’s Lemma Derivation of Black-Scholes Stock Pricing Model Solving Black-Scholes Recall our stochastic dierential equation to model stock prices: dS = σ dX + µ dt S where µ is known as the asset ’s drift ‚ a measure of the average rate
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to a camp. If 15% of the children in the school are first graders‚ and the 18 children are selected at random from among all 6 grades at the school‚ find the mean and variance of the number of first graders chosen? The mean is 2.7‚ and the variance is 2.3. n = 18 p = .15 q = .85 Mean = np = 18 x .15 = 2.7 Variance = npq = (18 x .15 x .85) = 2.295 ≈ 2.3 3. A producer plans an outdoor regatta for May 3. The cost of the regatta is $8000‚ which includes advertising‚ security‚ printing
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following is a random sample of six entertainment expenses (dinner costs for four people) from expense reports submitted by members of the sales force. $157‚ $132‚ $109‚ $145‚ $125‚ $139. Calculate the mean and sample variance(s^2) and standard deviation. Mean = 807/6 = 134.5. Sample Variance = (109925 – (807^2/6)/6-1 = (109925 – 108541)/5 = 1384/5 = 276.8. Standard Deviation = √276.8 = 16.6373. ***the 109925 is all values of x individually squared and then summed together. ***the 6-1 is because it
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