Is it Binomial
Let's assume 11% of all FSU students have a Discover card. We randomly select 200 students and count the number of those students who have a Discover card.
Yes
Binary: Have or don't have a Discover card
Independent: Random sample of students
fixed Number of trials: Picked 200 students
Same probability: Each pick has 11% probability
A recent survey found that 24% of homes in Tallahassee have a pool. Randomly select 20 houses and count the number of them that have pools.
n =
p =
n = 20
p = 24%
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the expected/mean number of students who will go to Starbucks?
Expected value = n x p = 30 x 14% = 4.2
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the probability that of the 30 students selected, 3 of them will go to Starbucks?
P(X = 3) =
Binompdf(30, 0.14, 3) = 18.98%
Randomly sample 500 FSU students and record which year they are in.
No.
Binary: No it is not binary, the outcomes can be Freshman, Sophomore, Junior, Senior, or Graduate
A recent survey found that 24% of homes in Tallahassee have a pool. Randomly select 20 houses and count the number of them that have pools.
What is the sample space
S = {0, 1, 2, ..., 20}
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the variance of the number of selected students who will go to Starbucks?
Var(X) = n x p x (1-p) = 30 x 14% x (1-14%) = 3.6
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the probability that 10 or fewer of the students will go the Starbucks?
Binomcdf(30,0.14,10) = 99.83%
Select a random group of 5 students who are walking together on campus and are all friends. Count the number of those students who have attended a football game this season.
No
Independent: Groups of students walking together probably have similar personalities and game attendeces.
A recent survey found that 24% of homes in Tallahassee have a pool. Randomly select 20 houses and count the number of them that have pools.
What is the shape of the distribution?
p < 50%
So it will be right skewed
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the standard deviation of the number of selected students who will go to Starbucks?
Std(X)
= sqrt(Var(X))
= sqrt(n x p x (1-p))
= sqrt(30 x 0.14 x (1-0.14))
= 1.9
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the probability that 5 or more of the students will go to Starbucks?
P(X >= 5)
= 1 - P(X <= 4)
= 1 - Binomcdf(30,0.14,4)
= 1 - 58.7%
= 41.3%
Walk around campus and randomly pick students until you find 15 students who are Freshman.
No
fixed Number of trials: Could end up asking anywhere from 15 to an infinite number of students.
A recent survey found that 24% of homes in Tallahassee have a pool. You go out and randomly select 20 houses and count the number of them that have pools.
Your friend also goes out and randomly selects 10 houses and counts the number of them that have pools.
You add up your number of houses with pools and their number of houses with pools.
What is the distribution of the total number of houses with pools?
X ~ Binomial(n = 20, p = 0.24)
Y ~ Binomial(n = 10, p = 0.24)
X + Y ~ Binomial(n = 30, p = 0.24)
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
How many students do you expect to not go to Starbucks.
Not go to Starbucks ~ Binomial(n = 30, p = 0.86)
Expected value = n x p = 30 * 86% = 25.8
or
Expect Not To Go = n - Expect to Go = 30 - 4.2 = 25.8
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the probability that 1 or 2 of the 30 students will go to Starbucks?
P(X = 1 or X = 2)
= P(X=1) + P(X=2)
= Binompdf(30,0.14,1) + Binompdf(30,0.14,2)
= 6.4% + 18.9%
= 25.3%
Randomly select 100 FSU students and count the number of them that got a ticket to the last football game.
No
Same probability: Each student has a different probability of getting a ticket based on their year.
A recent survey found that 24% of homes in Tallahassee have a pool.
Another survey found that 58% of homes in Gainesville have a pool.
Pick 20 random homes from each city and record the number of them that have pools.
Which city will have a more symmetric distribution of the number of selected homes with pools?
Gainesville.
The closer p is to 50%, the more symmetric the distribution will be.
A recent FSU survey found that 14% of all students go to Starbucks in any given day. A UCF survey found that 35% of all students go to Starbucks in any given day.
Randomly select 30 students from each school and count the number of them that will go to Starbucks on that day.
Which school has the higher variance of the number of students that will go to Starbucks?
X = FSU, Y = UCF
Var(X) = n x p x (1-p) = 30 x 14% x (1-14%) = 3.6
Var(Y) = n x p x (1-p) = 30 x 35% x (1-35%) = 6.8
UCF will have the higher variance.
A recent FSU survey found that 14% of all students go to Starbucks in any given day.
Randomly select 30 students and count the number of them that will go to Starbucks on that day.
What is the probability that between 2 and 8 selected students will go to Starbucks?
P(2 <= X <= 8)
= P(X <= 8) - P(X<=1)
= Binomcdf(30,0.14,8) - Binomcdf(30,0.14,1)
= 98.2% - 6.4%
= 91.8%