Question

Part I: Chi-squared Distribution & the Central Limit Theorem

The idea here is that you will explore a type of rv on your own (we only briefly mentioned this one in class)

a) Imagine sampling 50 values from χ²(8) a chi-squared distribution with 8 degrees of freedom.
According to the CLT (central limit theorem), what should be the expected value (mean) of this sample? You should not need to do any coding to answer this.

This is worth 1/2 EC point.

Part I: Chi-squared Distribution & the Central Limit Theorem

b) Imagine sampling 50 values from χ²(8) a chi-squared distribution with 8 degrees of freedom.

What should be the variance of this sample? You should use the CLT (central limit theorem) and should not need to do any coding to answer this.

This is worth 1/2 EC point.

II. Numeric probability approximations

The idea here is that you can use numeric data to approximate a probability instead of calculating it mathematically.

The life time of a flashlight battery (in hours) is exponentially distributed with λ = 1/120. Two such batteries (each in a different flashlight) are operating independently. Use R commands to answer the following questions.

b) Using your simulated vectors of each battery's lifetime from part a) estimate the probability that the both lasers each last more than 75 hours.

Report the answer to the nearest 4 digits.

This is worth 1 EC point.

Answer 1

PPFA Code for II

X=rexp(500,1/120) #Simulated life of battery 1 with 500 observations
Y=rexp(600,1/120) #Simulated life of battery 2 with 600 observations
x75=X>75 #Boolean vector, gives true if X(i) is greater than 75
y75=Y>75 #Boolean vector, gives true if Y(i) is greater than 75
p1=sum(x75)/500 #Proportion of battery 1 life lasting more than 75 hours
p2=sum(y75)/600 #Proportion of battery 2 life lasting more than 75 hours
p=p1*p2 #Probability that both the battery lasts more than 75 hours

We got p1 = 0.54

p2=0.5017

and P =0.2759