Calculus in Medicine Calculus in Medicine Calculus is the mathematical study of changes (Definition). Calculus is also used as a method of calculation of highly systematic methods that treat problems through specialized notations such as those used in differential and integral calculus. Calculus is used on a variety of levels such as the field of banking‚ data analysis‚ and as I will explain‚ in the field of medicine. Medicine is defined as the science and/or practice of the prevention
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DDM CASE STUDY CHAPTER – 5 HAPPY BULLS AND WORRIED BEARS * OVERVIEW OF THE CASE * End run provides two schemes: 1. Worried bear 2. Happy Bulls * With EndRun’s Worried bear fund scheme you can earn 400% rate of return in times of recession. * With EndRun’s Happy Bulls fund scheme you can earn 12 times your initial
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MATH133 Unit 5: Exponential and Logarithmic Functions Individual Project Assignment: Version 2A Show all of your work details for these calculations. Please review this Web site to see how to type mathematics using the keyboard symbols. IMPORTANT: See Question 1 in Problem 2 below for special IP instructions. This is mandatory. Problem 1: Photic Zone Light entering water in a pond‚ lake‚ sea‚ or ocean will be absorbed or scattered by the particles in the water and its intensity‚ I‚ will be attenuated
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CASE STUDY: 1 The bulbs manufactured by a company gave a mean life of 3000 hours with standard deviation of 400 hours. If a bulb is selected at random‚ what is the probability it will have a mean life less than 2000 hours? Question: 1) Calculate the probability. 2) In what situation does one need probability theory? 3) Define the concept of sample space‚ sample points and events in context of probability theory. 4) What is the difference between objective and subjective probability
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CHAPTER 1 SECTION 1: WHAT IS STATISTICS? MULTIPLE CHOICE 1. You take a random sample of 100 students at your university and find that their average GPA is 3.1. If you use this information to help you estimate the average GPA for all students at your university‚ then you are doing what branch of statistics? |a. |Descriptive statistics | |b. |Inferential statistics
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Deterministic techniques assume that no uncertain exists in model parameters. A: True An inspector correctly identifies 90% of the time. For the next 10 products‚ the probability that he makes fewer than 2 incorrect inspections is .736. A: Use Binomial table to discover ‚ add 3 probabilities for 0‚1‚2 A continuous random variable may assume only integer values within a given interval. A: False A decision tree is a diagram consisting of circles decision nodes‚ square probability nodes and branches
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1. In a pen factory there is a small chance 1/500 for any pen to be defective. The pens are supplied in packs of 10. Use this probability to calculate the approximate number of packets containing no defective‚ one defective and two defective pens‚ respectively in a consignment of 20‚000 packets [ e^(--0.02) =0.9802 ] Ans. : 19604‚ 392‚ 3.92=4 respectively 2. A manufacturer who produces medicine bottles finds that 0.1% of the bottles are defective. The bottles are packed in the boxes
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L={116*(Y/Yn) for (Y/Yn)>0.008856 903.3*(Y/Yn); Otherwise a = 500*(f(X/Xn –f(Y-Yn)) b = 200*(f(X/Xn –f(Z-Zn)) where Xn ‚Yn and Zn are the tristimulus values of the reference white. A median filter is applied on the L band in order to preserve edges and to reduce noise . A Contrast Limited Adaptive Histogram Equalization (CLAHE) technique is used. To enhance the contrast and the separability between exudates and the background ‚ a Contrast Limited Adaptive
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Figure 12.4: e – log sv’ plot 5. Recompression Index. Determine it from the plot of voids ratio (e) vs log σ plot as (Figure 12.4 or 12.5) 12.11 6. Coefficient of compressibility. It is calculated as follows av = 0.435 Cc/Avg. pressure for the increment 12.12 where Cc = compression index av = Coefficient of compressibility
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Table 2A and 3B 3A) we performed A priori chi-square test for these two tables. The main reason for the chi-square is to find out if the expected value is any different from the observed value. This part of the observation tested the null hypothesis‚ which states that‚ if when Ear#1(which was test-crossed kernel) is counted and approves of the Mendelian expectation of 1:1:1:1 phenotypic ratio. The chi-square for this data was excepted to accept the null hypothesis. The reason that it was expected
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