COFFEE USE - HEART RATE RESEARCH PROJECT COFFEE USE - HEART RATE Part 1 The study used for this project is a nursing and health sciences study. One particular healthcare facility is trying to determine if it is a good idea to provide coffee in the waiting room for the patients. There are several other facilities that serve tea‚ coffee‚ and water‚ so this health care facility wants to determine if there is sufficient evidence to show that coffee increases the patient heart rates. The question
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Mean | 474.0075 | Mean | 454.2225 | Mean | 106 | Median | 437 | Median | 417.5 | Median | 96 | Maximum | 975 | Maximum | 975 | Maximum | 282 | Minimum | 169.9 | Minimum | 165 | Minimum | 28 | Standard Deviation | 197.29003 | Standard Deviation | 192.5177534 | Standard Deviation | 58.2168207 | Based on the chart‚ the mean was calculated by adding up the sum of the list and divide 40‚ which the number of the total listed prices. The mean is 474‚007.5‚ which mean the average of the listed
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(monthly) return‚ the standard deviation of return and the coefficient of correlation between the returns. Using Excel‚ produce a chart showing alternative risk-return combinations in terms of 5% (e.g. 0% in WBK and 100% in WES‚ 5% in WBK and 95% in WES‚ 10% in WBK and 90% in WES‚ etc.) from the two investments. Interpret your results‚ also in comparison to investing in either of the individual stocks. Meanwhile please also calculate the minimum standard deviation of the portfolio contains these
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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 rates
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scores be used to compare scores from different distributions? 2. For the following set of scores‚ fill in the cells. The mean is 74.13 and the standard deviation is 9.98. Raw score Z score 68.0 ? ? –1.6 82.0 ? ? 1.8 69.0 ? ? –0.5 85.0 ? ? 1.7 72.0 ? 3. Questions 3a through 3d are based on a distribution of scores with and the standard deviation = 6.38. Draw a small picture to help you see what is required. a. What is the probability of a score falling between a raw score of 70 and 80? b. What
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iii) University Book Store Student Computer Purchase Program 1. Background. The Busy Biker is a bike shop located close to Tokyo University and owned by Nomura Hideki. He began the shop 20 years ago and sell new and second hand bicycle after fixing it too. The Business grew and demand continued to increase. Nomura must determine how many bicycles he need to have in stock at every beginning of semester. Below here is the data that help Nomura to forecast and to know how accurate
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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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= 712.44 standard deviation = square root of 712.44 = 26.69% coefficient of variation= CV= σ/ = 26.69%/11.40% = 2.34 expected return = = P1r1+P2r2+P3r3 expected return market= .3*15%+.4*9%+.3*18%= 13.5% expected return stock j= .3*20%+.4*5%+.3*12%=11.6% standard deviation =( r1- )2*P1+( r2- )2*P2+( r3- )2*P3 then square rooted standard deviation of market = ((15% - 13.5%)2*.3 + (9% - 13.5%)2*.4 + (18% -13.5%)2*.3) then square rooted = 3.85% standard deviation of stock j = ((20%
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manager receives a daily salary of $200 plus $85 per car sold. What is the expected value and standard deviation of her salary? (Hint: Use the ‘Summary of the Laws of Expected Value and Variance’ slide in the lecture notes.) Exercise 5.4. Now assume the company managing the network gets penalised $900 for each interruption. Calculate the expected daily penalty. Calculate the standard deviation of the daily penalty. (Hint: Use the ‘Summary of the Laws of Expected Value and Variance’ slide in the
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to verify the answer. Check and see the effect on acceptance of pins‚ when the mean and standard deviations are manipulated. Identify the most profitable situation based on cost of reengineering. 1. What is the probability that a batch will be acceptable to the consumer and if the probability is large enough to be an acceptable level of performance? 2. If the population mean and standard deviation of the length of the pins are adjusted in order to improve the percentage accepted‚ which one do you
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