Actuarial Outpost Hypothetical Exam question
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 Short-Term Actuarial Math Old Exam C Forum

#1
09-30-2016, 10:18 AM
 rgreenlee Member Non-Actuary Join Date: Jun 2014 Location: in a van down by the river Posts: 1,075
Hypothetical Exam question

I took this exam awhile back, but I ran across something that I wanted to check...it reminded me of a problem on the easier scale of questions one might run into on this exam.

What is the 95% confidence interval of the average of 14 independent uniformly (0,1) distributed numbers?

Is it (.34878,.65122)

(sketch: st dev of the average = .077151675, 1.96 standard deviations away from mean)
#2
09-30-2016, 10:53 AM
 bamaguy Member CAS SOA Join Date: Jul 2015 Studying for S College: University of Alabama Favorite beer: Ballast Point - Big Eye Posts: 164

I believe that checks out, yeah
#3
09-30-2016, 02:24 PM
 BuddhaWilliams Member SOA Join Date: Mar 2016 Posts: 152

I got the same answer as well
#4
10-02-2016, 12:00 AM
 StayCoolKD Member CAS Join Date: Sep 2014 College: Alumni University of Washington Posts: 272

U(0,1) --- mean=1/2 variance=1/12
14 independent--- mean=14/2 var=14/12
ave--- mean/n var/n

I don't get it anyone can explain the solution? thanks

"ave of 14 independents"
#5
10-02-2016, 01:44 AM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

$\displaystyle \text{Var}\left(\dfrac{\sum_{i=1}^{14}X_{i}}{14} \right)=\text{Var}\left(\frac{1}{14} \cdot \sum_{i=1}^{14}X_{i} \right)=(\frac{1}{14})^{2} \cdot \text{Var}\left(\sum_{i=1}^{14}X_{i} \right)=\frac{1}{14^2} \cdot \sum_{i=1}^{14}\text{Var}\left(X_{i} \right)=\frac{1}{14^2} \cdot \sum_{i=1}^{14}\text{Var}(X)=\frac{1}{14^2} \cdot 14\text{Var}(X)=\dfrac{\text{Var}(X)}{14}$
#6
10-02-2016, 04:52 AM
 Demo99 Member SOA Join Date: Jan 2013 Location: CA Posts: 234

Quote:
 Originally Posted by Z3ta $\displaystyle \text{Var}\left(\dfrac{\sum_{i=1}^{14}X_{i}}{14} \right)=\text{Var}\left(\frac{1}{14} \cdot \sum_{i=1}^{14}X_{i} \right)=(\frac{1}{14})^{2} \cdot \text{Var}\left(\sum_{i=1}^{14}X_{i} \right)=\frac{1}{14^2} \cdot \sum_{i=1}^{14}\text{Var}\left(X_{i} \right)=\frac{1}{14^2} \cdot \sum_{i=1}^{14}\text{Var}(X)=\frac{1}{14^2} \cdot 14\text{Var}(X)=\dfrac{\text{Var}(X)}{14}$
Well put. For this exam I'd recommend just knowing that VAR(X_bar)=VAR(X)/n
#7
10-02-2016, 01:48 PM
 StayCoolKD Member CAS Join Date: Sep 2014 College: Alumni University of Washington Posts: 272

Quote:
 Originally Posted by StayCoolKD U(0,1) --- mean=1/2 variance=1/12 14 independent--- mean=14/2 var=14/12 ave--- mean/n var/n I don't get it anyone can explain the solution? thanks "ave of 14 independents"
ok thanks! I made mistake on the expected value: E(X_bar)=E[X]
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