CO2 the owner of a apple plantation, plans to replace the brand of fertilizer that they currently use if the harvest for the current month is lower than 58 tons/ha. It is known that his farm’s monthly yield follows the normal distribution with mean 65 tons/ha and variance 25 tons2/ha2. What is the probability that the brand of fertilizer currently used will be replaced? What would be the maximum yield of his lowest 15% monthly harvests? Use the table given for the z value

Suppose you would like to keep a test tube at 48F (exactly), but your refrigeration unit exhibits some random fluctuations, so that the difference between the temperature at which you set the thermostat and the true temperature is distributed normally with mean 0 and variance 1.96. What is the highest you can set the thermostat and be 95% sure that the true temperature is no more than 48F?

disp hp qsec wt Use the following visualization to answer questions 2 - 4. mpg drat drat mpg qsec wt hp disp -1 -0.8 -0.6-0.4 -0.20 0.2 0.4 0.6 0.8 1

Pax World Balanced is a highly respected, socially responsible mutual fund of stocks and bonds. Vanguard Balanced Index is another highly regarded fund that represents the entire U.S. stock and bond market (an index fund). The mean and standard deviation of annualized percent returns are shown below. The annualized mean and standard deviation are for a recent 10-years period.†.   If x represents return and s represents risk, then explain why the coefficient of variation can be taken to represent risk per unit of return. From this point of view, which fund appears to be better? Explain. A. Since the CV is s/s2 we can say that the CV represents the risk per unit of return; the Pax fund appears to be better because the CV is smaller. B. Since the CV is s/s2 we can say that the CV represents the risk per unit of return; the Vanguard fund appears to be better because the CV is smaller.     C. Since the CV is s/x we can say that the CV represents the risk per unit of return; the Pax fund…

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.97.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 2424 samples is 8.18.1 ppm with a variance of 0.250.25. Does the data support the claim at the 0.10.1 level? Assume the population distribution is approximately normal. Step 3 of 5:  Specify if the test is one-tailed or two-tailed.

Based on predictive modeling, this upcoming Winter is going to have very unpredictable weather. The National Weather service is considering abolishing specific long-term forecasts, and instead will utilize a statistical model that predicts a mean of six significant snowfalls per month. Use this information to answer the following question correctly: Trying to be forward-thinking, I'm going to try planning my one month long Winter Break based around this model's predictions. I will only travel for the holidays if there is at most one significant snowfall over winter break. What is the probability of this happening? O 0.0174 O 0.0149 O 0.0025 O 0.0542

Suppose that the annual rate of return for a common biotechnology stock is normally distributed with a mean of 5% and a standard deviation of 6%. Find the probability that the one-year return of this stock will be positive. Round your answer to at least four decimal places.

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 24 samples is 8.1 ppm with a variance of 0.25. Does the data support the claim at the 0.1 level? Assume the population distribution is approximately normal. Step 5 of 5:  Make the decision to reject or fail to reject the null hypothesis.

Please let me know if I calculated relative risk correctly. Thank you for your assistance.

Assume the probability that a maple tree at age 10 grows less than 110 cm is equal to 0.4. If the height of maple trees at age 10 are estimated to be normally distributed with mean u cm and variance 100 cm. find u.