The results for all 10 grocery surveys used in this lesson were generated using a mathematical simulation, which you can find on the Simulation tab of the spreadsheet. To use it, enter a population mean and population standard deviation in the two yellow cells at the top of the worksheet and follow the instructions. Every time a new set of data is generated, the data changes, providing a new sample. You can copy this data and paste it into the data plot tool to analyze it. (You don’t have to do that in this activity, but it’s there for you to experiment with, if you like.) Any given set of sample data generated in this way would not typically have the same mean, distribution, and standard deviation as the population values, but the simulated sample data would be consistent with a randomly selected sample from such a population.

Note that the simulated population mean is 1.73, and the simulated standard deviation is 0.657. Imagine that this random sample survey was simulated an infinite number of times. In that case, the population mean () of all the samples would be $1.73. All the sample means would be normally distributed, according to the central limit theorem.

Part A
If the standard deviation of the population is $0.657 (as it is in the simulation), what is the standard error of the mean? Round up your answer to the nearest tenth of a cent.

Cone w has a radius of 8 cm and a height of 5 cm. square pyramid x has the same base area and height as cone w. paul and manuel disagree on how the volumes of cone w and square pyramid x are related. examine their arguments. which statement explains whose argument is correct and why? paul manuel the volume of square pyramid x is equal to the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is three times the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is (area of base)(h) = (200.96)(5) = 1,004.8 cm3. paul's argument is correct; manuel used the incorrect formula to find the volume of square pyramid x. paul's argument is correct; manuel used the incorrect base area to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect formula to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect base area to find the volume of square pyramid x.