lab homework 3
Lab Homework
#save your plots as an image file and upload separately (export > save as image)
#save them as png with filename your_name_Lab3_plot
#Exercise 1
#The rain probability for the month of July in the city of Portland is 72%
#(a) What is the probability that it rains exactly 11 days
#(b) What is the probability that it rains 11 or fewer days?
#(c) What is the probability that it rains 27 or more days?
#(d) What is the probability that it rains more than 20 but less than 25 days?
#Exercise 2
#Suppose the winnings of gamblers at Las Vegas are normally distributed with
#mean $670 and standard deviation $38.
#(a) What is the probability of winning exactly $500?
#(b) What is the probability of winning $500 or less?
#(c) What is the probability of winning more than $800?
#(d) What is the probability of winning between $500 to $875?
#(e) Generate 10000 random numbers using rnorm function. Show how this
#distribution would look like. Add in the plot the mean, median and 20%
#trimmed mean (provide these values).
#Exercise 3
#In the previous problem, how much someone has to win to be in the top 5%?
——————————————————————-
LAB 3 EXAMPLE
#Lab 3-Contents
#1. Binomial Probability Distribution
#2. Normal Probability Distribution
#————————————–
# 1. Binomial Probability Distribution
#————————————–
# The binomial distribution is one of the most important distributions in statisitcs.
# Like it’s name suggests, bi-nomial, indicates that we are dealing with
# numbers that have only two possibilites (eg. Yes/No, Dead/Alive etc.)
#There are three main types of problems dealing with binomial probabilites:
#A) Finding the probability of EXACTLY x successes (Yes responses, deaths, boys)
#B) Finding the probability of < x successes OR > x successes
#C) Finding the probability of < x1 successes AND > x2 successes
# Let’s try one by semi-hand (we’ll have R do some computations)
# and then learn how to use R to solve these kinds of problems.
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXAMPLE 1: Let’s say the probability for divorce is 45%. Out of 20 married couples,
#what is the probability of EXACTLY 5 of them getting divorced?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
N=20 #n
K=5 # x
p=0.45
q=1-p
Nf=factorial(N) #n!
Kf=factorial(K) # x!
Nf/(Kf*factorial(N-K)) * p^K * (1-p)^(N-K)
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#Binomial Probability (EXACT): dbinom(k, n, p)
#Where k: number of successes, n: number of trials,
#p: probability of success
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#Let’s use the dbinom() command from R to answer this question
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXERCISE 1-1: Use the dbinom() command to solve the problem in Example 1
#
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
dbinom(5, 20, 0.45)
# What if we wanted to know the probability of 5 OR fewer divorces?
# One strategey is to compute the probability for 5, 4, 3, 2, 1, 0 divorces and sum the probabilities
p5=dbinom(5, 20, 0.45)
p4=dbinom(4, 20, 0.45)
p3=dbinom(3, 20, 0.45)
p2=dbinom(2, 20, 0.45)
p1=dbinom(1, 20, 0.45)
p0=dbinom(0, 20, 0.45)
p0+p1+p2+p3+p4+p5
# This is a bit tedious, so let’s just use a new command called pbinom()
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#Binomial Probability (x <= k): pbinom(k, n, p)
#Where k: <= number of successes, n: number of trials,
#p: probability of success
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXERCISE 1-2: Use the pbinom() command to solve the problem of computing the probability for
# 5 or fewer divorces.
#
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
pbinom(5, 20, 0.45)
#??????????????????????????????????????????????????????????????#
#Thought Question 1: How would the command change if the question
# was worded as “Find the probabilty for less than 5 divorces”?
pbinom(4, 20, 0.45)
#??????????????????????????????????????????????????????????????#
# The pbinom() command is useful, but only gives us probabilities for <= k
# What if we wanted to know the probability of having >= k responses or > k?
# Because the probabilities have to add to exactly 1, we can use some simple math to figure this out
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXAMPLE 2: The probability for divorce is 45%. Out of 20 married couples,
#what is the probability of >= 6 of them getting divorced?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
# We know from above that the probability of <= 5 couples divorcing is 0.05533419
# If we think about it, then the probability of >=6 is just 1-0.05533419 = 0.9446658
# We could even use the pbinom() command with this by putting the 1 minus in front
1-pbinom(5, 20, 0.45)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXERCISE 1-3: Use the pbinom() command to solve the problem of computing the probability for
# > 7 divorces.
#
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
1-pbinom(7, 20, 0.45)
# Thus far we’ve covered how to answer 2 out of the 3 possible binomial probability probelms (A and B).
# Let’s think about how we could answer a question asking about a range of possible sucesses.
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXAMPLE 3: The probability for divorce is 45%. Out of 20 married couples,
#what is the probability of >= 7 and <= 10 divorces?
# 7=< p<=10 ?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
# This problem has 2 parts.
#1) Compute the probability for <=10 divorces
pbinom(10, 20, 0.45)
#2) Compute the probability for <= 6 divorces (YES, 6 not 7)
pbinom(6, 20, 0.45)
# The difference between these two probabilites is the probability between 7 and 10
pbinom(10, 20, 0.45) – pbinom(6, 20, 0.45)
#Note: Probability will always be positive, so make sure you do the subtraction in the correct order
#??????????????????????????????????????????????????????????????#
#Thought Question 2: How would computing the probability
# in example 3 change if the direction of the signs were reversed?
# eg. <=7 & >=10
#p<=7 and p>=10
#??????????????????????????????????????????????????????????????#
pbinom(7, 20, 0.45)
1-pbinom(9, 20, 0.45)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXERCISE 1-4: Use the pbinom() command to solve the problem of computing the probability for
# > 4 AND <= 15 divorces. 4<p<=15
#
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
pbinom(15, 20, 0.45) – pbinom(4, 20, 0.45)
#————————————–
# 2. The Normal Probability Distribution
#————————————–
# Probabilities associated with continous variables are often computed using the normal distribution
# Rather than taking about the number of successes in a given number of trials, now we are concerned with
# first defining the properties of the normal distribution and determine the probabilites of having certain
# values along that distribution
#Let’s look at the standard normal distribution
# The standard normal has a mean of 0 and SD of 1
#I’m going to create a random variable with 10,000 observations of mean 0 and sd=1
set.seed(1)
x=rnorm(10000, mean=0, sd=1)
mean(x); sd(x)
hist(x)
#In the standard normal distribution, we can compute (or look up in a table) what the probability values
# should be for a Z score less than some value.
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXAMPLE 4: What is the probability of having < -1 (called a Z score) on the standard normal distribution
#
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
# First off, we can actually look this up in the appendix of your textbook.
# Page 300, beings the listing of the Z scores and their probabilities.
# From the textbook, this value is: 0.1587
# This means that the probability of having < -1 is 0.1587
#Let’s look at this on a plot:
plot(density(x))
abline(v= -1, col=”red”)
#Instead of the textbook, we could use the pnorm() function in R
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#Normal Dist Probability (Z <= z): pnorm(Z, mean, sd)
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
pnorm(-1, 0, 1)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXERCISE 2-1: Thinking about what we learned from using the binomial functions, how could we use
# pnorm() to find the probability of having a Z score > 1.96 on the standard normal distribution (mean=0, sd=1)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
1-pnorm(1.96, 0, 1)
#We can generalize this to normal distributions that do not have a mean of 0 and SD of 1
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXAMPLE 5: What is the probability of having a bmi < 25.2 if the mean bmi is 35 and SD = 5
#
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#We can use pnorm() in this situation as well
#1)
pnorm(25.2, 35, 5)
# 2)
Z=(25.2-35)/5
pnorm(Z,0,1)
#Notice that the answer we get: 0.0249979 is the same as the answer we got in Exercise 2-1.
#??????????????????????????????????????????????????????????????#
#Thought Question 3: Why do you think the answers are the same in
# example 5 and exercise 2-1?
#??????????????????????????????????????????????????????????????#
(25.2-35)/5
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXERCISE 2-2: The average age for incoming college freshman is 18.2 years with SD=0.5.
# What is the probability that a freshman will be between 18 and 19 years old?
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
pnorm(19, 18.2, .5) – pnorm(18, 18.2, .5)
# We can also do these problems in reverse, where we are given the probability and have to find the values
# at which that probability exists. When we do this, we are calculating the quantile of the normal distribution.
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
#Normal Dist Probability (Quantile): qnorm(p, mean, sd)
#^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^#
# Let’s think about the problem we did in example 5
# We found the probability of a bmi < 25.2 when the mean of the distribution was 35, SD=5
# The probability was 0.0249979
# Let’s see if qnorm() returns the correct result:
qnorm(0.0249979, 35, 5)
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
#EXERCISE 2-3: The average age for incoming college freshman is 18.2 years with SD=0.5.
# A researcher tells you that the 65% of the freshman were less than…
# He can’t seem to remember what age it was at that 65% of the freshman were less than, help him out.
#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#*#
qnorm(0.65, 18.2, .5)
#????????????????????????????????????????????????????????????????????????????????????#
#Thought Question 4: What if instead of knowing the age for the bottom 65%
# we wanted to know how old a person needed to be in the top 10% of age in their class
#?????????????????????????????????????????????????????????????????????????????????????????#
qnorm(0.9, 18.2, .5)
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