Data collected at elementary s

Data collected at elementary schools in DeKalb County, GAsuggest that each year roughly 25% of students miss exactly one dayof school, 14% miss 2 days, and 26% miss 3 or more days due tosickness.

Solution:

a) the probability that a student chosen at random doesn’t missany days of school due to sickness :

probability does not miss = P(X = 0)

P(X = 0) =1-0.25 – 0.14 – 0.26

=0.35

b) the probability that a student chosen at random misses nomore than one day :

P(X≤1) = P(X=0) + P(X=1)

= 0.35 + 0.25

= 0.60

c) the probability that a student chosen at random misses atleast one day:

P(X≥1) = 1- P(X=0)

= 1- 0.35

= 0.65

d) If a parent has two kids at an elementary school, theprobability that neither kid will miss any school :

Assume that events are independent,

Therefore, Probability that neither kid miss any school

= 0.35 * 0.35

= 0.1225

e) If a parent has two kids at an elementary school, theprobability that both kids will miss some school, ie at least oneday:

probabilty that both kids miss some school

= (1 – 0.35)*(1 – 0.35)

= 0.65 * 0.65

= 0.4225

f) Do you think it was reasonable to assume the probabilitiesare independent in parts (d) and (e)? The kids are siblings, and ifone gets sick it probably … the chance that the other one getssick. So whether or not one misses school due to sickness isprobably … of the other

we assumed that going to school probability for both kids isindependent of each other, That does not seem much reasonable asbeing kids of same parent their school going tendency should bepositivily correlated.