Transcript 2.12 - Open Online Courses

EXERCISE 2.12 – SCHOOL ABSENCES
Background:
Data collected at elementary schools in DeKalb County, GA suggest that
each year roughly 25% of students miss exactly one day of school,
15% miss 2 days, and 28% miss 3 or more days due to sickness.
QUESTIONS
(a) What is the probability that a student chosen at random doesn't miss any days of
school due to sickness this year?
(b) What is the probability that a student chosen at random misses no more than one
day?
(c) What is the probability that a student chosen at random misses at least one day?
(d) If a parent has two kids at a DeKalb County elementary school, what is the
probability that neither kid will miss any school? Note any assumption you must
make to answer this question.
(e) If a parent has two kids at a DeKalb County elementary school, what is the
probability that both kids will miss some school, i.e. at least one day? Note any
assumption you make.
(f) If you made an assumption in part (d) or (e), do you think it was reasonable? If you
didn’t make any assumptions, double check your earlier answers.
ANSWER – A
First, we convert the percentages that we are given into decimals:
•
25% is converted to 0.25
•
15% is converted to 0.15
•
28% is converted to 0.28
We can now calculate the probability that the student didn’t miss any days of school
by subtracting the probability that they missed one or more days from 1.
The equation will be set up as follows:
1 – ((the probability that the student only missed 1 day) + (the probability that the
student missed 2 days) + (the probability that the student missed 3 or more
days))
In mathematical terms it would be:
1 – ((0.25) + (0.15) + (0.28)) = 0.32
ANSWER - B
To calculate the likelihood that a student missed no more than one day we
perform a function similar to Answer A. Except, in this case we want to
know whether they missed two or more days.
Set the equation up as follows:
1 – ((the probability that the student missed 2 days) + (the probability that
the student missed 3 or more days)
In mathematical terms:
1 – ((0.15) + (0.28)) = 0.57
ANSWER – C
This question asks us the probability that a student misses at least
one day. We are given the probability of students missing one
day, two days, or three or more. Therefore, in order to answer the
question we simply need to add those probabilities together.
Solution:
 0.25 + 0.15 + 0.28 = 0.68
ANSWER – D
This question asks what the probability the neither child misses any school.
We apply the multiplication rule for independent processes.
P(A and B) = P(A) x P(B)
In problem A we determined the probability that a random student doesn’t
miss any days of school to be 0.32. We apply this to our current
question
P(ChildA and ChildB) = P(ChildA) x P(ChildB)
=0.32 x 0.32
= 0.1024
ANSWER – E
This is a similar problem to D and we apply the same logic. In
problem C we determined the probability of a randomly selected
student missing at least on day to be 0.68.
Again, we apply the multiplication rule for independent processes to
calculate our answer. It should look like this:
P(ChildA and ChildB) = P(ChildA) x P(ChildB)
= 0.68 x 0.68
= 0.4624
LETS GRAPH!
Student Absences
30
Percent of Students Missed
25
20
15
10
5
0
1
2
Days Missed
3