In a certain Algebra 2 class of 29 students, 7 of them play basketball and 14 of them
1 answer:
Answer:
<em>Two possible answers below</em>
Step-by-step explanation:
<u>Probability and Sets</u>
We are given two sets: Students that play basketball and students that play baseball.
It's given there are 29 students in certain Algebra 2 class, 10 of which don't play any of the mentioned sports.
This leaves only 29-10=19 players of either baseball, basketball, or both sports. If one student is randomly selected, then the propability that they play basketball or baseball is:
P = 0.66
Note: if we are to calculate the probability to choose one student who plays only one of the sports, then we proceed as follows:
We also know 7 students play basketball and 14 play baseball. Since 14+7 =21, the difference of 21-19=2 students corresponds to those who play both sports.
Thus, there 19-2=17 students who play only one of the sports. The probability is:
P = 0.59
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Answer:
a)
Operating room x 1 2 3 4
P(x) 0.15 0.25 0.4 0.2
b)
The graph of probability distribution is in attached image.
c)
All the probabilities lies in the range (0 to 1) and the probabilities add up to 1 so, the computed probability distribution is the valid probability distribution.
Step-by-step explanation:
Number of operating rooms Days
1 3
2 5
3 8
4 4
a)
Sum of frequencies=1+2+3+4=10
Operating rooms x frequency f Relative frequency
1 3 3/20=0.15
2 5 5/20=0.25
3 8 8/20=0.4
4 4 4/20=0.2
So, using the relative frequency approach, a discrete probability distribution for number of operating rooms in use on any given day is
Operating room x 1 2 3 4
P(x) 0.15 0.25 0.4 0.2
b)
The graph of probability distribution is in attached image.
c)
There are two conditions for a probability distribution to be valid
1. All probability must ranges from 0 to 1.
2. All probabilities must add up to 1.
We can see that all the probabilities lies in the range (0 to 1), so, condition 1 is satisfied.
For condition 2,
sum[p(x)]= 0.15+0.25+0.4+0.2=0.4+0.6=1.
As the probabilities add up to 1, so the condition 2 is also satisfied.
Thus, the computed probability distribution is the valid probability distribution.
