Normal Distribution

2703 WordsJan 24, 201211 Pages
Final Exam Review Questions Solutions Guide You will probably want to PRINT THIS so you can carefully check your answers. Be sure to ask your instructor if you have questions about any of the solutions given below. 1. Explain the difference between a population and a sample. In which of these is it important to distinguish between the two in order to use the correct formula? mean; median; mode; range; quartiles; variance; standard deviation. Solution: A sample is a subset of a population. A population consists of every member of a particular group of interest. The variance and the standard deviation require that we know whether we have a sample or a population. 2. The following numbers represent the weights in pounds of six 7year old…show more content…
Explain mathematically. Solution: (a) The total number of transfer students is 270. The total number of students in the survey is 500. P(Transfer) = 270/500 = .54 (b) The total number of part time students is 210. The total number of students in the survey is 500. P(Part Time) = 210/500 = .42 (c) From the table we see that there are 100 students which are both transfer and part time. This is out of 500 students in the sample. P(transfer ∩ part time) = 100/500 = .20 (d) This is conditional probability and so we must change the denominator to the total of what has already happened. There are 100 students which are both transfer and part time. There are 210 part time students. P(transfer | part time) = 100/210 ≈ .4762 (e) P(part time | transfer) = 100/270 ≈ .3704 (f) The definition of independent is P(A|B) = P(A). To test we ask if P(part time | transfer) = P(part time)? Is .3704 = .42? No, there for the events are not independent. We could also test P(transfer | part time) = P(transfer). Is .4762 = .54? Again, the answer is no. (g) For events to be mutually exclusive their intersection must be 0. In part c we found that P(transfer ∩ part time) = 100/500 = .20. Therefore the events are not mutually exclusive. 11. A shipment of 40 television sets contains 3 defective units. How many ways can a vending company can buy five of these units and receive no defective units? Solution: There are 37 sets which are not defective. There are
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