Distributions.Rmd

---
title: "problem_set_3"
author: "Lydia Holley"
date: "2024-01-24"
output: pdf_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)

```

# Uniform distribution

Create a datatset drawn from a uniform distribution with 10,000 samples taken between the values of 4 and 6

```{r uniform}
# create uniform distribution
unif_dist<-runif(10000,4,6)

```

## Question 1

Visualize that distribution with a histogram

```{r histogram}
# produce histogram
hist(unif_dist)

```

# Binomial Distribution

If two mice who are carriers for albinism (but not albino themselves) mate, their offspring have a 1/4 probability of being albino. Based on this information conduct the following analyses

```{r binomial}
# set probability of a 'success'
prob_alb = 0.25

```

## Question 2

What is the probability that exactly 2 offspring in a litter of 10 are albino?\
- 0.2815676

```{r question2}
# find probability for Q 2
dbinom(2,10,prob_alb)

```

## Question 3

Plot the probability distribution function for albinism in a litter of 10 mice.

```{r prob dist}
# plot probability distribution function
x<-1:10
plot(x,dbinom(x,size=10,prob=.25),type="h")

```

## Question 4

What is the probability that there are 8 or more albino mice in a litter of 10 mice?\
- 0.000415802

```{r question4}
# probability of 8 or more
# find probability of less than 8 and subtract from 1

q4<-1-pbinom(7,10,prob_alb)
print(q4)
```

## Question 5

What is the probability of at least 3 but no more than 5 albino mice in a litter of 10\
- 0.4546795

```{r question5}
# find probability for getting 3, 4, or 5 albino mice
# can either find each individual and add or subtract less than 3 from 5 or less

q5<-pbinom(5,10,prob_alb) - pbinom(2,10,prob_alb)
print(q5)

```

# Multinomial Distribution

## Question 6

If the parental birds shown above have 20 offspring throughout their lifetime, what is the probability of getting equal numbers of Yellow/Normal, Yellow/Clear, White/Normal, and White/Clear offspring?\
- 3.383959e-05

```{r question6}
# equeal numbers would be 5 of each
dmultinom(x=c(5,5,5,5),prob=c(0.5625,0.1875,0.1875,0.0625))

```

## Question 7

The White/Clear birds are the most valuable for your research. How many offspring should you plan for so that there is a \>80% chance of having at least 8 White/Clear baby birds?\
- You should plan for 181 or more offspring

```{r question7}
# more than 80% chance of having 8 or more white/clear
# loop until the probability of getting 8 is more than 80%
# print the outcomes. don't need multinomial since we only care about one of the outcomes

chances <- 0
x = 0
while (chances <= 0.8) {
  x = x + 1
  chances <- pbinom(8,x,0.0625,lower.tail = FALSE)
  
}

print(chances)
print(x)
```

# Standard Normal Distribution

```{r standard normal dist}
# set the rules of the test for our investigation
n_questions<-100
error_rate<-0.12
correct_answer<-0.88

```

## Question 8

Take a sample of 31 student scores from the binomial distribution of a 100 question where the rate of success is random and 88%. What is the mean score in that classroom?\
- 89.06452

```{r question8}
# take sample of 31 from 100 questions with p of 0.88 (should be around 88)
classroom<-rbinom(31,100,0.88)
mean_score<-mean(classroom)
print(mean_score)
```

## Question 9

There are 120 classes in the school district and they are all 31 students. Simulate 120 classrooms taking the exam and then plot the mean classroom score in a histogram.\

```{r question9}
# use loop to simulate 120 classes of 31 students
classes<-vector()
numClasses<-120

for (i in 1:numClasses){
  class_score<-rbinom(31,100,0.88)
  class_mean<-mean(class_score)
  classes[i]<-class_mean
}

hist(classes,breaks=25)

```

## Question 10

You note that this distribution looks like a normal distribution! From your sample of 120 mean classroom scores across the district, calculate the mean and standard deviation of the average classroom scores (from the 31 students.)\
- mean of 88.03441 and a standard deviation of 0.6294889

```{r question 10}
# find mean and standard deviation
q10_mean<-mean(classes)
q10_sd<-sd(classes)
print(q10_mean)
print(q10_sd)

```

## Question 11

Use the mean and standard deviation of the mean classroom scores across the district to plot a normal distribution of scores under the assumption that all scores are due only to random error.\

```{r question11}
# plot normal distribution using the mean and standard deviation
x<-seq(75,100,length=100)
plot(x,dnorm(x, mean=q10_mean,sd=q10_sd), type="h")

```

## Question 12

Based on your model, you examine the two lowest scoring classrooms. Classroom A had a mean exam score of 87.5. Classroom B had a mean exam score of 80.4. Which classroom(s) could have obtained this mean score due simply to the error of the exam?\
- Classroom A based on my model.

# Chi-Squared Distribution

## Question 13

Chi-Squared Value at which 95% of values are below this threshold for

-   2 Degrees of Freedom
    -   5.991465
-   6 Degrees of Freedom
    -   12.59159
-   10 Degrees of Freedom
    -   18.30704
-   100 Degrees of Freedom
    -   124.3421

```{r question13}
# find chi square values for which 95% of values fall beneath
qchisq(0.95,2)
qchisq(0.95,6)
qchisq(0.95,10)
qchisq(0.95,100)

```

# T-Distribution

## Question 14

Plot the T-distribution for 50 classes (degrees of freedom) where the distribution is centered over the score 88.\

```{r question14}
# plot t distribution centered over 88 with 50 df
curve(dt(x-88,49), from=75, to=100 )
```

## Question 15

Based on your model, you examine the two lowest scoring classrooms in this school. Classroom A had a mean exam score of 87.5. Classroom B had a mean exam score of 80.4. Which classroom(s) could have obtained this mean score due simply to the error of the exam?\
- Classroom A could have obtained this mean score simply due to error of the exam

# F-Distribution

## Question 16

Which degrees of freedom do you use in the numerator to ensure the test is the most conservative?\
- I would use df 30 in the numerator to ensure test is conservative.

```{r question16}
# compare the f values for which 95% of values fall below depending on df
pf(0.95, (12-1), (31-1))
pf(0.95, (31-1), (12-1))
```