Q1. Because cheating has been on the increase, the Rector has set up a special investigation. As part of
this, a private investigation company has been engaged, and they arrange for one of their agents to
"register" for a particular statistics course. The agent finds out that a group of 10 students met at a bar
and bought a copy of the mid-term quiz and its solution in advance. Unfortunately for them, the
professor who wrote the exam made a rather unusual mistake in the solution to question 2 of the quiz,
so it is pretty easy to spot who was cheating if their answer matches this mistake. There are 65 students
registered for the course.
4 a) [ 4 ] If the marker checks 5 papers as a sample, what is the chance none of them belongs to one of
the cheaters? Be sure to name any methods or models you use and show how you get your answer.
Hypergeometric
N,S,n,K :
65 10 5 0
P(K=0| N, S, n) =
C(10, 0) C(55, 5) / C(65, 5)=
1*3478761 / 8259888 = 0.4212
1 Hypergeom, 1 params, 1 setup, 1 answer.
Also possible by
55/65 * 54/64 * 53/63 * 52/62 * 51/61=
0.421163216740953
b) In a separate investigation, professors belonging to the activist group Cascade Rounding Angers
Professors want to correct the improper practice of "Cascade Rounding" where 1.23456 is rounded to
1.24 by first rounding to 1.2346, then 1.235, and finally 1.24. (THIS IS WRONG, for those who do it
The correct answer is 1.23). First, they want to know how many students do this, so set up a calculation
where the mean of seventeen numbers is $1.23456,
then ask for it to be rounded to the nearest cent.
The believe fully 12% of students are CR users.
4 1b1)
[ 2 ]
If they sample 7 papers from a very large collection of exams, what is the probability no
student cascade rounds? Be sure to name any methods or models you use and show how you get your
answer.
Binomial
(1)
n=7, p=.12
(1)
P(K=0 | n, p) =
(1-p)^n
=
0.40867559636992
1 + 1
marks
3 1b2) [ 3 ] If they sample 7 papers from a very large collection of exams, what is the probability at
least 2 students of the seven cascade rounds?
1 - P(0) - P(1) =
1 - 0.40867559636992
- 0.39009943289856
= 0.20122497073152
1 setup, 1 parts, 1 answer
5 1c) [ 5 ] A nationwide study is conducted on cascade rounding and there are many thousands of high
school graduates asked to round 1.23456 to 2 decimals. What is the probability that a random sample of
198 such graduates returns less than 12 "cascade rounders"?
P(K<12 | n=198, p=.12) ~= P(z < (11.5 - np) /sqrt(npq) = P(z <
-2.68117965919122 )
but
use z = -2.68
to get
0.00368110800917498
or 0.37 % probability
1 setup, 1 CC, 1 z, 1 lookup, 1 answer
Bonus Question:
2 1d) [ 2 ] Suppose you had conducted the study in ( c ) only for the U of Ottawa where 546 students
were in the population. What happens to your answer in ( c )? Specify your answer with the new value
of the resulting
probability.
Must reduce std deviation of "K" by using FPC, so z gets bigger.