Descriptive Statistic Inferential Statistics Parameter vs Statistics Variable a. Categorical Statistic estimates Parameter b. Quantitative estimates ‚ sample mean ‚ population i. Discrete mean s‚ sample standard estimates ‚ population ii. Continuous deviation standard deviation Random Variable estimates P‚ population ˆ p ‚ sample Sampling Distributions proportion proportion Parameter (Defines a population) Statistic (calculated from sample to estimate a parameter) Central Limit
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P1-2010-2011-Spring Bus531 Standard Normaal Probability Problems 1- A university reported admission statistics for 3339 students who were admitted for freshmen for the fall semester of 1998. Of these students‚ 1590 had taken the Scholastic Aptitude Test (SAT). Assume the SAT verbal scores were normally distributed with a mean of 530 and a standard deviation of 70. a – What percentage of students were admitted with SAT verbal scores between 500 and 600? b - What percentage of students were
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2/8 (Friday) Standard Deviation cont’d Example: Amount of money earned by new immigrants Sample: 12‚ 15‚ 16‚ 20‚ 25‚ 36‚ 40 Step 1: Find the mean * x̄=164/7 * mean is 23.4 Step 2: Find how much each observation deviates from the mean * 12 - 23.4 = -11.4 * 15 – 23.4 = -8.4 * 16 – 23.4 = -7.4 * 20 – 23.4 = -3.4 * 25 – 23.4 = 1.6 * 36 – 23.4 = 12.6 * 40 – 23.4 = 16.6 * Note: all observations below mean will be negative‚ all above will be positive Step
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in this experiment. When trialing the octagon parachute‚ the results show very accurate data. It shows that it is the slowest speed‚ and experiences the most air resistance. The square parachute has the second most accurate data‚ based on the standard deviation from the mean. The speed of the square is the second slowest‚ and experiences great air resistance. The circle experiences fast rates placing it in third. The data is not as precise as the square and octagon. The rectangle experiences very fast
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1880. What are the advantages and disadvantages in using these data to help estimate the expected rate of return on U.S. stocks over the coming year? If we assume that the distribution of returns remains reasonably stable (same expectation and standard
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964 Security A has an expected rate of return 6%‚ a standard deviation of returns of 30%‚ a correlation coefficient with the market of -.25‚ and a beta coefficient of -0.5. Security B has an expected return of 11% a standard deviation of returns of 10%‚ a correlation with the market of .75 and a beta coefficient 0.5. Which security is more risky? Why? From the problem‚ the standard deviation of Security A is 30% and the standard deviation of Security B is 10%. Now let’s look at the CV of each
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Planning Introduction Stomata are pores on the bottom of leaves of vascular plants. The function of stomata is to allow gaseous exchange (transpiration) between the plant and the external environment (see diagrams below). This process is carried out by the plant‚ which controls the opening and closing of stomata via the guard cells. The more stomata a leaf contains the greater the rate of gas exchange and the less water it will be able to accumulate. Stomatal pores are formed between
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economic‚ political‚ tax and market conditions. Standard market risk factors are stock prices‚ interest rates‚ foreign exchange rates‚ and commodity prices. 2.2 Definition of Standard Deviation In statistics and probability theory‚ standard deviation shows how much variation or "dispersion" exists from the average (mean‚ or expected value). A low standard deviation indicates that the data points tend to be very close to the mean‚ whereas high standard deviation indicates that the data points are spread
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Introduction to Management Science‚ 10e (Taylor) Chapter 11 Probability and Statistics 1) Deterministic techniques assume that no uncertainty exists in model parameters. Answer: TRUE Diff: 1 Page Ref: 489 Main Heading: Types of Probability Key words: deterministic techniques 2) Probabilistic techniques assume that no uncertainty exists in model parameters. Answer: FALSE Diff: 1 Page Ref: 489 Main Heading: Types of Probability Key words: probabilistic techniques 3) Objective
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20 0.15 0.10 0.10 Find P( X < 3) A.0.20 B.0.15 C.0.80 D.0.85 Given that Z is a standard normal random variable‚ P(-1.0 < Z < 1.5) is A.0.9332 B.0.0919 C.0.8413 D.0.7745 The standard deviation of a probability distribution must be: A.a negative number B.a nonnegative number C.less than the value of the mean D.a number between 0 and 1 If Z is a standard normal random variable‚ then the value z for which P(-z < Z < z) equals
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