Question2 i. Define a discrete randomvariable and its probability distribution.
ID: 2915789 • Letter: Q
Question
Question2
i. Define a discrete randomvariable and its probability distribution.
ii. What are the two basicproperties of all probability distributions?
iii. Consider the population ofeight students whose ages are given in the table. Let Xdenote the age of a randomly selected student.
18
22
22
18
21
27
22
21
a) Compute the population mean andthe population standard deviation.
b) Construct a probabilitydistribution of X.
c) Find the expected value as wellas the standard deviation for the random variable X.
18
22
22
18
21
27
22
21
Explanation / Answer
i. A discrete random variable is a random variable that is definedover a discrete set - as in, there are only a finite number ofoutcomes (like two possible outcomes if you flip a coin, comparedto infinite outcomes if you measure someone's height). The probability distribution is a function that gives a probabilityof occurence to each of those possible outcomes. One example of a discrete random variable is the number of typos ina page of text. Assuming that it is equally likely for a typo tohappen at any letter, this random variable is poisson distributed,with a mean that is dependent on how likely it is for a typo tohappen. ii. Two basic properties of probability distributions are, 1. The sum of all probabilities is one. This is because thereis a 100% chance that you will get some kind of outcome (even ifthe outcome is that nothing has happened). 2. Following from #1, no event has probability greater thanone. iii. a. The population mean is just the average of the eightnumbers, 1/8*(18*2+22*3+2*21+27)=21.375. The standarddeviation is a measure of how spread out the data is from the meanvalue. To compute it, subtract the mean from each data point,square the result, than average these eight numbers (one from eachdata point). Then take a square root. In this case you shouldget 2.64. b. The probability distribution is : normal(21.375,2.64^2) which is the normal distribution with themean as computed above, and variance = square of the standarddeviation computed above. c. Since we used the computed mean and standard deviation to definethis distribution, the expected values are the same as the computedvalues from the sample.
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