In this project, students will demonstrate their understanding of probability and the normal and binomial distributions.
IQ scores are approximately normally distributed with: X ∼ N(μ=100,σ=15)
#Proportion of population with an IQ greater than 65
pnorm(q=65, mean=100, sd=15, lower.tail=FALSE)
## [1] 0.9901847
99% of the population has an IQ greater than 65.
#IQ correlated with top 5% of the population
qnorm (p = 0.95, mean = 100, sd = 15)
## [1] 124.6728
An IQ score of about 124.7 represents the top 5% of the population, or, the 95th percentile. This means that someone with an IQ score of 124.7 has scored higher than 95% of test takers.
Recall our definition: A value is considered unusual if it lies more than two standard deviations from the mean.
#IQ values that mark lower and upper bounds of "usual" range
100-30
## [1] 70
100+30
## [1] 130
#Proportion of population outside of lower and upper bounds
pnorm(q= 70, mean=100, sd=15) + pnorm(q=130, mean=100, sd=15, lower.tail=FALSE)
## [1] 0.04550026
Two students took different standardized tests.
Alex took the SAT and scored 1650. Taylor took the ACT and scored 27.
Assume the distributions:
SAT∼N(1500,300) ACT∼N(21,5)
#Alex
(1650-1500)/300
## [1] 0.5
#Taylor
(27-21)/5
## [1] 1.2
Taylor performed better relative to other ACT test takers because her Z score is higher.
Comparing raw scales alone would be misleading because the SAT and the ACT have different grading scales. This means they also have different means and standard deviations. Although it looks like Alex has a higher score, Taylor has a better score relative to other ACT test takers because her Z score is higher.
You are taking a 15-question multiple choice quiz and each question has 5 options (a,b,c,d) and you randomly guess every question.
#Average correct answers
15*1/5
## [1] 3
#Probability every question is correct
dbinom(x=15, size=15, prob=1/5)
## [1] 3.2768e-11
#Probability every question is incorrect
dbinom(x=0, size=15, prob=1/5)
## [1] 0.03518437
#Probability exactly 10 questions are correct
dbinom(x=10, size=15, prob=1/5)
## [1] 0.000100764
#Probability of 10 or more questions being correct
pbinom(q=9, size=15, prob=1/5, lower.tail = FALSE)
## [1] 0.0001132257
No, I do not believe this claim. The probability of that happening is extremely small, about 0.0001. This makes it extremely unlikely the student was randomly guessing on the quiz.
#Scoring 12 or more correct
pbinom(q=11, size=15, prob=1/5, lower.tail = FALSE)
## [1] 1.011253e-06
A company schedules 10 employees for a shift. Each employee independently shows up with probability: p = 0.85
Let X = number of employees who show up
The company needs at least 8 workers to operate normally.
#Probability that less than 8 employees show up
pbinom(q=7, size=10, prob=0.85)
## [1] 0.1798035
#Probability of enough workers showing up
pbinom(q=7, size=10, prob=0.85, lower.tail = FALSE)
## [1] 0.8201965
There is an 82% chance the company has enough workers staffed and an 18% chance they do not. The probability of having enough workers is higher, but there’s still a significant chance they may not.
Yes, management should schedule more than 10 employees to increase their chance of having enough workers. With each employee there is a 85% chance they’ll show up, adding more employees increases the probability of having enough workers each shift.
ACT scores are approximately normally distributed where: X ∼ N(21,5) a. Use R to simulate 10,000 ACT scores.
# 10,000 ACT score simulation
ACT <- rnorm(n = 10000, mean = 21, sd = 5)
#ACT scores above 30
prop.table(table(ACT>30))
##
## FALSE TRUE
## 0.9636 0.0364
#Theoretical ACT scores above 30
pnorm(q = 30, mean = 21, sd = 5, lower.tail = FALSE)
## [1] 0.03593032
The two values are similar because they’re both normally distributing the same situation. They’re not the same because the simulation pulls from a random sample of ACT scores so there can be slight differences.
Create your own real-world situation that could be modeled using either a binomial distribution or a normal distribution.
Your problem must include: * A description of the situation * Identification of reasonable parameters (mean, sd OR n, p) * One probability calculation in R * A written interpretation of the result
Examples might include: * basketball free throws * weather events * exam scores * products being defective
The Uniform Bar Exam is a test taken before you can practice law. UBE scores are approximately normally distributed with: X ∼ N(279,33)
What proportion of test takers will score a 300 or higher?
#Proportion of test takers with a 300 or higher
pnorm(q = 300, mean = 279, sd = 33, lower.tail = FALSE)
## [1] 0.2622697
About 26% of test takers should score a 300 or higher on the UBE.