A small daycare center for young children has only 3, 4, 5, 6, or 7 children in
ID: 3024286 • Letter: A
Question
A small daycare center for young children has only 3, 4, 5, 6, or 7 children in attendance on any one day. On half of the days there are three children in attendance and on a quarter of the days there are 4 children in attendance. The probability on 5 and on 6 children in attendance is 0.10 each, with the remainder of the probability on the event that 7 children will attend.
What is the variable?
What are the values of the variable?
What is the probability distribution of the variable?
What is the expected number of children who will attend the small daycare center?
If an adult must be present for every 3 children, what is the probability that the daycare center will need 3 adults?
If an adult must be present for every 3 children, what is the probability that the daycare center will need 2 adults?
If an adult must be present for every 3 children, what is the probability that at least 2 adults are needed?
In an adult must be present for every 3 children, what is the probability that only one adult is needed?
Explanation / Answer
(1)here variable is number of children in day care centre and
(2) its value are 3,4,5,6 and 7
(3) Probability distriburion
p(3)=0.5,p(4)=0.25,p(5)=0.1,p(6)=0.1 and p(7)=0.05 other wise zero
(4) expected number of childeren=3*0.5+4*0.25+5*0.1+6*0.1+7*0.05=1.5+1+0.5+0.6+0.35=4.45
(5)for 3 adults there will be need of 9 children so probability that the daycare center will need 3 adults will be 0 , because variable 9 is out of range. its range is 3 to 7
(6)probability of 2 adults= probability of more than 3 children but less than 6 children=p(4)+p(5)+p(6)=0.25+0.1+0.1=0.45
(7) probability that at least 2 adults=probability of one adults+probability of two adults=p(3)+p(4)+p(5)+p(6)=0.5+0.25+0.1+0.1=0.95
(8) probability of only one adult=probability of 3 children=p(3)=0.5