Here is data from a college university on the numbers of courses a student is en
ID: 3170338 • Letter: H
Question
Here is data from a college university on the numbers of courses a student is enrolled in. Use the following information to answer the next set of questions. In this example, x is the number of classes a student is enrolled in. There are 15,000 students in this sample. Is this variable discrete or continuous? Calculate the probability mass function (pmf) and the cumulative distribution function (cdf). Calculate the E[X] of the data. Let's assume that each class is 3 credits. And, we transformed the data from number of classes a student was enrolled, to the number of hours a student was enrolled in. Therefore, the new distribution is h(x) = 3x. Calculate E[h(x)] and V[h(x)]. Use, the E[X] you calculated and V[X] = 1.2651.Explanation / Answer
a. The discrete variable refer to a variable, values of which are countable, and continuous variable refer to a variable, whose values are measured. The number of courses/classes a student is enrolled in countable and finite. Therefore, x is a discrete random variable.
b. Probability mass function refer to a probability distribution function, involving only discrete values of x. The properties are as follows:
0<=P(X=x)<=1, Sigma P(X=x)=1
The cumulative distribution function refer to the probability a random variable, X takes on a value less than or equal to some particular value of a. This is expressed as F(a)=p(X<=a)
[definitions have been adapted from Harnett,Hayes]
p(x) 0.01 (150/15,000) 0.03 (450/15,000) 0.13 0.25 0.39 0.17 0.02
F(x) 0.01 0.04 (0.01+0.03) 0.17 0.42 0.81 0.98 1.00
c. E[x]=Sigma i=1 to k Pi*xi
=0.01*1+0.03*2+0.13*3+0.25*4+0.39*5+0.17*6+0.02*7
=4.57 (ans)
d. E[h(x)]=E[3x]=3E[x] [By applying rule, E(aX)=aE(X)]
=3*4.57
=13.71 (ans)
V[h(x)]=V[3x] [By applying rule, Var(aX)=a^2Var(X)]
=9V[x]
=9*1.2651
=11.3859 (ans)
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