**3.12 School Absences** Data collected at elementary schools in DeKalb County, GA suggests that each year roughly 25% of students miss exactly one day of school due to sickness, 15% miss 2 days, and 28% miss 3 or more days. **Questions:** (a) What is the probability that a randomly chosen student misses at least one day? (b) What is the probability that a randomly chosen student misses at least 2 days? (c) What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year? (d) What is the probability that a student chosen at random misses no more than one day? (e) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss school? Note any assumption you make. (f) 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 assumptions. If you made an assumption in part (e) or (f), do you think it was reasonable? If you didn’t make any assumptions, double check your earlier and then return to this question. **Solution Steps:** - To solve part (a), calculate the probability of missing at least one day: \(0.25 + 0.15 + 0.28 = 0.68\). - To solve part (b), calculate the probability of missing at least two days: \(0.15 + 0.28 = 0.43\). - To solve part (c), find the probability of a student not missing any days: \(1 - 0.68 = 0.32\). - To solve part (d), find the probability of a student missing no more than one day: \(0.25 + (1 - 0.68) = 0.25 + 0.32 = 0.57\). For parts (e) and (f), consider independent probabilities and assumptions such as similarity in chances of missing school between siblings.
**3.12 School Absences** Data collected at elementary schools in DeKalb County, GA suggests that each year roughly 25% of students miss exactly one day of school due to sickness, 15% miss 2 days, and 28% miss 3 or more days. **Questions:** (a) What is the probability that a randomly chosen student misses at least one day? (b) What is the probability that a randomly chosen student misses at least 2 days? (c) What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year? (d) What is the probability that a student chosen at random misses no more than one day? (e) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss school? Note any assumption you make. (f) 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 assumptions. If you made an assumption in part (e) or (f), do you think it was reasonable? If you didn’t make any assumptions, double check your earlier and then return to this question. **Solution Steps:** - To solve part (a), calculate the probability of missing at least one day: \(0.25 + 0.15 + 0.28 = 0.68\). - To solve part (b), calculate the probability of missing at least two days: \(0.15 + 0.28 = 0.43\). - To solve part (c), find the probability of a student not missing any days: \(1 - 0.68 = 0.32\). - To solve part (d), find the probability of a student missing no more than one day: \(0.25 + (1 - 0.68) = 0.25 + 0.32 = 0.57\). For parts (e) and (f), consider independent probabilities and assumptions such as similarity in chances of missing school between siblings.
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Problem 1P
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Transcribed Image Text:**3.12 School Absences**
Data collected at elementary schools in DeKalb County, GA suggests that each year roughly 25% of students miss exactly one day of school due to sickness, 15% miss 2 days, and 28% miss 3 or more days.
**Questions:**
(a) What is the probability that a randomly chosen student misses at least one day?
(b) What is the probability that a randomly chosen student misses at least 2 days?
(c) What is the probability that a student chosen at random doesn’t miss any days of school due to sickness this year?
(d) What is the probability that a student chosen at random misses no more than one day?
(e) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss school? Note any assumption you make.
(f) 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 assumptions. If you made an assumption in part (e) or (f), do you think it was reasonable? If you didn’t make any assumptions, double check your earlier and then return to this question.
**Solution Steps:**
- To solve part (a), calculate the probability of missing at least one day: \(0.25 + 0.15 + 0.28 = 0.68\).
- To solve part (b), calculate the probability of missing at least two days: \(0.15 + 0.28 = 0.43\).
- To solve part (c), find the probability of a student not missing any days: \(1 - 0.68 = 0.32\).
- To solve part (d), find the probability of a student missing no more than one day: \(0.25 + (1 - 0.68) = 0.25 + 0.32 = 0.57\).
For parts (e) and (f), consider independent probabilities and assumptions such as similarity in chances of missing school between siblings.
Expert Solution
Step 1
(a)
Based on the given information, the following information is known:
P(Students miss exactly one day of school)=0.25(=25/100)
P(Miss 2 days)=0.15(=15/100)
P(Miss 3 or more days due to sickness)=0.28(=28/100)
The probability that a student chosen at random does not miss any days of school due to sickness this year is given below:
P(Does not miss any days of school due to sickness)=1− {P(Students miss exactly one day of school)+P(Miss 2 days)+P(Miss 3 or more days due to sickness)}
=1−{0.25+0.15+0.28}
=1−0.68=0.32
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