1. Shari carlos Mark shawky Nate stocks
Probability
Shari carlos
Mark shawky
Nate stocks

2. 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.

3. 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.

4. 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

5. 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

6. 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:
= 0.68

7. 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 We apply this to our current question P(ChildA and ChildB) = P(ChildA) x P(ChildB) =0.32 x 0.32 =
The multiplication rule for independent processes can be found on page 78

8. 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 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 =
Use the multiplication rule for independent processes it can be found on page 78

9. Lets Graph!