Problem 1

Question) We often use the Gaussian distribution when all we know about a process is the mean and the variance. Suppose that a random variable X > 0 has mean = 10 and standarddeviation = 6. Is it appropriate to use a Gaussian distribution for X? Why or why not?

10

Answer) We know the random variable X has been defined to take values X > 0. The mean is close to 0 with a relatively large standard deviation. If we approximate X as a Gaussian distribution we would have roughly 5% of the values less than 0, which would not correctly reproduce X . This is not a great approximation but if it is all you know to use then it could be tolerable in some applica-tions. Although perhaps a Gamma distribution modeled for X would work better.

Problem 2

Treat the population density of catalyst sites as a Gaussian distribution.

rDE_ := 2 p sDE2 -1

2 Exp- DE-DE02

2sDE2

;

The assumed reaction rate has the form

kDE_ := k0 Exp- DE

R T;

The domain of activation energies is assumed to be DE œ [-¶, ¶] (even though in reality there are no negativeactivation energies)

Lets first check to see if the distribution is normalized.

-¶

¶

rDE „DE = 1

Yes the distribution is normalized.

Part A) What is the average activation energy, DE?

E E E

DE = DE0

Part B) What is the average rate, k?

kE E E

k = k0 „sDE

2

2 R2 T2-DE0

R T

Part C The population density of sites is approximated

as Gaussian. We want to find the maximum contributer to the rate.

The integrand is kDE rDE. We want to find a maximum for

the product of these two functions wrt. activation energy.

Set the derivative of the integrand = 0 and solve for the extremized activation energy.

2 hw4-questions1and2.nb

SolveDkDE rDE, DE ã 0, DE Simplify

DE Ø DE0 -sDE

2

R T

DE = DE0 -sDE

2

RT

The product of k[DE] r[DE] is a maximum at the point of intersection for the schematic shown below.

DE= DE0 -sDE

2

RT

r[DE]

k[DE]

2 4 6 8 10

0.2

0.4

0.6

0.8

1.0

Part D Only a small fraction of sites have the probability of exisiting with a

reasonably low activation energy. The rate depends negative exponentially on the activation

energy. So even though the number of sites becomes increasingly small toward the left-

tail end of the Gaussian distribution the rate exponentially increases.

hw4-questions1and2.nb 3

Problem 3

μx 1.833358

σx 1.503571 N=100

μy 1.762671

σy 1.865165 NOTE: Sample standard deviations need to be divided by

μxy 5.595008 sqrt(1/(N‐1)) not sqrt(1/(N))

σxy 2.363401

ρxy 0.842746

The correlation and covariance are nonzero when the variables 

are approximatly linearly related.

There is a positive correlation between variables X and Y from looking at the plot below. 

This matches the sign of my calculated parameters. 

It is a good idea to check your reported numbers for data consistency.

‐1

0

1

2

3

4

5

6

y

‐2

‐1

0

1

‐4 ‐2 0 2 4 6x