Bio 205 Lab W/8:00 Enzyme II Write-Up Methods: My partner and I ran two experiments to measure the activity of the enzyme horseradish peroxidase under varied conditions. The first of which measured the effects of altered pH levels‚ while the goal of the second was to examine the effects of varied temperatures. To test the effects of pH on horseradish peroxidase‚ we began by zeroing a Spec 20 with 5.0mL of substrate (25mM guiacol) at pH 6.5. Once the Spec 20 was accurately zeroed‚ we added 100μL
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the key issues are in regards to choice of technology‚ equipment and processes at the stage of formulation of Detailed Project Report? (10 Marks) 4. Given the activity mean and Standard Deviation‚ Find the probability that the project will take more than 10 weeks to complete. (20 Marks) Activity Mean Standard Deviation 1–2 2–3 1-3 5 4 8 1 1 1 5. For the following network data ‚ (20 Marks) (a) Identify the Critical Path and its duration (b) Calculate the total network slack time
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-------------------------------------------------------------------------------- 6 BASIC STATISTICAL TOOLS There are lies‚ damn lies‚ and statistics...... (Anon.) -------------------------------------------------------------------------------- 6.1 Introduction 6.2 Definitions 6.3 Basic Statistics 6.4 Statistical tests -------------------------------------------------------------------------------- 6.1 Introduction In the preceding chapters basic elements for the proper
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Charismatic Condition Mean 4.204081633 Standard Error 0.097501055 Median 4.2 Mode 4.8 Standard Deviation 0.682507382 Sample Variance 0.465816327 Kurtosis 5.335286065 Skewness -1.916441174 Range 3.5 Minimum 1.5 Maximum 5 Sum 206 Count 49 Confidence Level(95.0%) 0.196039006 In both the Charismatic and the punitive condition data sets there were 49 people surveyed. We know this because we were able to use descriptive statistics to show the count and that shows the number of people
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Statistics Course Project Part 1 Student: Zissis Koukoutaris Course : Statistics Professor: Dimitrios Fotiadis Semester: Fall-2012 Introduction This report is written by Koukoutaris Zissis a student from Northeastern University. Mr Fotiadis my professor of the MGSC 23021 assigned me the task. 1. Use methods of descriptive statistics to summarize the data. Comment on the findings 2. Develop estimated regression equations‚ first using annual income as the independent variable
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Name___________________________________ Provide an appropriate response. 1) Two random variables are normally distributed with the same mean. One has a standard deviation of 10 while the other has a standard deviation of 15. How will the graphs of the two variables differ and how will they be alike? 2) Which is larger‚ the area under the standard normal curve between -1 and 1‚ or the area under the standard normal curve between 0 and 2? Explain your reasoning. 3) Which of the variables below do you think will be roughly
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1. As the degrees of freedom increase‚ the t distribution approaches the b. normal distribution 2. If the margin of error in an interval estimate of μ is 4.6‚ the interval estimate equals b. [pic] 3. The t distribution is a family of similar probability distributions‚ with each individual distribution depending on a parameter known as the c. degrees of freedom 4. The probability that the interval estimation procedure will generate an interval that does
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for the variable Cost. But this does not account for the presence of variation in a sample that could affect the mean. We can use this value within a standard normal distribution of sample means to try and narrow down the true population mean within a confidence interval of probability. This is preferable because it takes into account the standard error that is intrinsic to all iterated samples of a population and uses that to capture the population mean (μ) within a range. It is not plausible to
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the 52nd percentile is .05 A normally distributed variable has a mean of 10 and a standard deviation of 2. The probability that a value between 7 and 9 is obtained is .2417 An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. The distribution of the lives approximated a normal distribution. The standard deviation was 1.2 hours. About 95.44% of the batteries failed between what two values? 16.6
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with a standard deviation of 3.6 percent (Exhibit 2). This is the only positive return compared to our proxy markets: the VFINX Index (-.4%)‚ NASDAQ 100 (-.7%)‚ and the VHGEX Index (-.1%). Along with generating the highest return‚ our portfolio had the highest risk. Compared to the 3.6 percent standard deviation of returns on our portfolio‚ the VFINX Index (1.87%)‚ the NASDAQ 100 (2.54%)‚ and the VHGEX Index (1.60%) all show less volatility (Exhibit 2 shows the standard deviations of the 4
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