UNSW Astrophysics Postgraduate Course
Module 3, Session 2
Optical and Infrared Techniques
Lecture 5: Data Reduction and Analysis
Tutorial Answers
http://www.phys.unsw.edu.au/ 
spd/pgc_astro/part3.html
1. (a) Binomial distribution, so expected number of failures 
.
(b) Probability that x students fail = 
.
So P( 
fail) = 1 - P(0 fail) - P(1 fail) - P(2 fail) - P(3 fail) -
P(4 fail)
= 1 - 0.0884 - 0.2228 - 0.270 - 0.2142 - 0.1223 = 0.0803.
ie 8% chance of 5 or more failing.
2. (a) Poisson distribution 
.
P(detecting 8 or more events in a day) = 1 - 
P(i events
seen)
= 1 - 0.1353 - 0.2707 - 0.2707 - 0.1804 - 0.0902 - 0.0301 -
0.0120 - 0.0034 = 0.0011.
(b) Considering just a 10 minute period then 
events expected. Do above sum using this value of 
and
you get 0 to about 20 decimal places!
3. (a) Binomial distribution with p= P(R) = 670/1000 and q = P(L) = 330/1000.
So uncertainty is 
.
Applies to both 
and 
.
(b) 
. 
. So 
.
(c) Assume A=0.400. So 
. Thus 0.4 = p - (1-p) so p = 0.7 and q = 1-p = 0.3. So
. Then 
.
4. The frequency shift 
.
So
, assuming the variables are independent.
We find that 
,
and
. So that 
. So 
. Note that if we were
measuring the actual frequency, instead of calculating what its shift would
be, then the uncertainty in its measurement is greater than the actual shift
in frequency!
5. Gaussian distribution. Let 
and look for
probability that 
. We seek 
.
From tables we find that 
so that 
, respectively.
6. Bin into 10 mark ranges. Expected number per bin is 
where 
is bin width and N total number
of observations, with 
. Events
within each bin are distributed according to the Poisson
distribution as we are extracting a random sample from a parent
population for that bin value. For the Poisson distribution
as 
. So the error estimate
per bin is simply 
.
Thus we can calculate 
, the observed number per bin, 
, the
expected number per bin and 
, their spread for that bin. We
find that 
. We have
10 bins so 9 degrees of freedom. ie 
. From tables 
and 
. Thus in repeated experiments the probability of
obtaining 
. This is a high
probability, so the data is consistent with coming from a Gaussian
parent population with 
and 
.
Anyone who answered 29 or 38 to the final part is failed for impertinence!
Worksheet for calculating the number of samples per bin, the expected number per bin, the uncertainity and the Chi-square statisitic from the deviations.
Comparison between binned data and best Gaussian fit, whose parameters
are determined from the mean and standard deviation of the data. The Chi-square
test says the data are consistent with having come from this parent distribution.