Math 1180 - Worksheet for Exam 3
Some Essential Formulas and Denitions
RV means random variable, C means for continuous RV, D means for discrete RV, CD means
for both continuous and discrete RVs.
Section 6.8: expectation (from 6.7, CD), median (CD), mode (CD), geometric mean (CD).
Section 6.9: range, percentile (CD), mean absolute deviation (CD), variance, standard deviation, coecient of variation
Section 7.1: often uses denition of conditional probability from section 6.4, joint distribution, marginal distribution, conditional distribution, conditional probability formula with
independent variables
Section 7.2: covariance, correlation (uses marginal distributions, expectation, covariance,
variance, standard deviation, and correlation formula)
Section 7.3: no independence requirement: E (X + Y ), E (aX ), var(X + Y ) using covariance, var(aX ). with independence requirement: E (XY ), var(X + Y ) without covariance,
cov(X; Y ) = 0.
Section 7.4: binomial distribution p.d., mean, variance, mode,
given to you on test)
n
k
(some will be on chart
Section 7.6: rst half: geometric distribution p.d., mean, variance, mode (or how to nd it),
median (or how to nd it), c.d. (or how to nd it), range
Problems
1. Find the expectation, median, mode, range, variance, standard deviation, and coecient of variation of the following p.d. and p.d.f.
X
j Probability
j
0:2
j
0:3
j
0:1
j
0:4
0
(a) 1
2
3
(b) f (x) = 1:5(x + x ) for 0:5 x 1:0. (Note the median cannot easily be found by
hand, but set up the problem.)
j Y = 1 j Y = 2 0:0 j 0:2 X = 0 j
2. Suppose X and Y have the following joint distribution: X = 1 j 0:2 j 0:1 X = 2 j
0:1 j 0:0 X = 3 j 0:1 j 0:3 (a) Find the correlation of X and Y . Do this by nding the marginal distributions
of X and Y (and note any similarity to the previous problem to possibly save
computations later). Then nd the expection, variance, and standard deviation of
each. Finally nd the covariance. Then you will have everything you need to nd
the correlation. Explain what this correlation implies.
(b) Find the cumulative distribution for X and the c.d. for Y .
(c) Are X and Y independent?
3. Suppose E (M ) = 1, E (N ) = 3, var(M ) = 0:5, var(N ) = 1, and cov(M; N ) = 0:1.
(a) Assume M and N are not independent. Find E (5M + 2N ), var(2N ), and var(M +
N ).
(b) Suppose M and N are independent (and change the value of the covariance accordingly). Find var(5M 2N ).
4. Suppose Q is a Bernoulli random variable with probability of success 0:3.
(a) How many times do you expect Q to equal 1 in 10 trials? What is the probability
that Q = 1 exactly three times? What is the standard deviation of Q? What is
the coecient of variation? How do you interpret the coecient of variation?
(b) When do you expect Q to rst equal 1 in discrete trials? What is the probability
that Q = 1 on the third trial? What is the standard deviation of Q? What is the
mode? What is the median? What is the coecient of variation? How do you
interpret the coecient of variation?
2
1.
Solutions
(a) E (X ) = 1:7, median= 1, mode= 3, range= 0 to 3, var(x) = 1:41, st.dev.= 1:187,
coe.var.= 0:698
(b) E (X ) = 0:789, median is solution to 0:75x + 0:5x = 0:5, mode= 1, range= 0:5 to
1:0, var(X ) = 0:0197, st.dev.= 0:14, coe.var.= 0:178.
(a) marginals: Pr(X = 0) = 0:2, Pr(X = 1) = 0:3, Pr(X = 2) = 0:1, Pr(X = 3) = 0:4
(note this matches 1a); Pr(Y = 1) = 0:4, Pr(Y = 2) = 0:6.
From 1a we know E (X ) = 1:7, var(X ) = 1:41, = 1:187.
E (Y ) = 1:6, var(Y ) = 0:24, = 0:49.
cov(X; Y ) = 0:2. corr(X; Y ) = 0:034. So when X is large, Y is small about
34% of the time.
(b) Pr(X 0) = 0:2, Pr(X 1) = 0:5, Pr(X 2) = 0:6, Pr(X 3) = 1:0.
Pr(Y 1) = 0:4, Pr(Y 2) = 1:0.
(c) No. One proof: Pr(X = 0 \ Y = 1) 6=Pr(X = 0)Pr(Y = 1).
(a) E (5M + 2N ) = 11, var(2N ) = 4, var(M + N ) = 1:7
(b) var(5M 2N ) = 16:5
(a) Binomial. E (Q) = 3, b(3; 10; 0:3) = 0:267, var(Q) = 2:1, = 1:45, coe.var.=0:48.
So values may sometimes be far from the expected value, but the expected value is
a decent predictor.
(b) Geometric. E (Q) = 3:3, Pr(Q = 3) = 0:147, = 2:789, mode= 1, median=
= 1:94, coe.var.= 0:845. So the values may be far from the expected value
and the expected value is not a great predictor.
2
2.
3
X
Y
3.
4.
ln(0:5)
ln(0:7)