Applied Statistics
Assignment 1
Due Fri 29 March 2013
Instructions
1. Cut-and-paste this document into word (or equivalent)
2. Show working where possible (eg cut-and-paste commands from an R ses- sion). 3. Submit your assignments on AUTonline
4. Remember that you can use the help system in R to provide hints.
5. Remember that you can try out ideas using the R system to check your thinking

Questions
1. (a) how many ways are there to choose 7 objects from 13? (5 marks)
> choose(13,7)
[1] 1716
> factorial(13)/(factorial(6)*factorial(7))
[1] 1716
(b) how many ways are there to arrange 12 objects? (5 marks)
> factorial(12)
[1] 479001600
> 1*2*3*4*5*6*7*8*9*10*11*12
[1] 479001600
(c) If a classroom has 25 children in it, how many ways are there to choose four children? (5 marks)
> choose(25,4)
[1] 12650
> factorial(25)/(factorial(21)*factorial(4))
[1] 12650
(d) If you have 30 children, then how many ways are there to put them into three classes of 10 each? (hint: Figure out how many ways are there to choose 10 children for the first class; and then multiply this by the number of ways of choosing 10 children for the second class) (10 marks)
> choose(30,10)*choose(20,10)
[1] 5.550997e+12
> (factorial(30)/(factorial(20)*factorial(10)))*(factorial(20)/(factorial(10)*factorial(10)))
[1] 5.550997e+12 (e) [harder] A manufacturer of widgets makes 10 widgets. Four com- panies wish to buy widgets; company A wishes to buy 4 widgets, company B wishes to buy 3 widgets, C buys 2 and company D buys one widget. How many ways can the manufacturer distribute the
10 widgets amongst the customers so each one recieves the correct number of widgets? (10 marks)
> choose(10,4)*choose(6,3)*choose(3,2)
[1] 12600
> (factorial(10)/(factorial(6)*factorial(4)))*(factorial(6)/(factorial(3)*factorial(3)))*(factorial(3)/(factorial(1)*factorial(2)))
[1] 12600
2. Here are some numbers: x mean(x)
[1] 10
> sum(x)/length(x)
[1] 10
(b) what is their mode? (5 marks)
> table(x) x 9 10 15 5 1 1
9 is their mode.
9 occur more often than any other number (five times).
(c) what is their median? (5 marks)
> median(x)
[1] 9
> sort(x)
[1] 9 9 9 9 9 10 15
> sort(x)[4]
[1] 9
(d) state why the mode is a poor representative value for this dataset
(10 marks)
9 is away from central tendency.
9 is at the extreme of the dataset.
(e) suppose, instead, that the numbers were x mean(x)
[1] 142857142865
> sum(x)/length(x)
[1] 142857142865
> median(x)
[1] 9
> sort(x)
[1] 9e+00 9e+00 9e+00 9e+00 9e+00 1e+01 1e+12
> sort(x)[4]
[1] 9
> table(x) x 9 10 1e+12 5 1 1
9 is mode.
Reason: because 1e12 is the highest number.
Mean is much larger because the mean is sensitive to the outlier of 1e12.
3. Suppose X is drawn from a standard normal distribution, mean 0 and standard deviation 1.
(a) What is the probability of X being less than 1.8? Give an R expres- sion and a numerical value to 7 signicant figures. (10 marks)
> pnorm(1.8)
[1] 0.9640697
> table(rnorm(1e6)1.9) to give and why? (5 marks)
Expect about 1e6*pnorm(1.9)[ie
Reason: We should expect about 971283 to be true. table(rnorm(1e6)>1.9) is close enough to our expectation.
> pnorm(1.9)
[1] 0.9712834
> 0.9712834*1e6
[1] 971283.4
> table(rnorm(1e6) dbinom(5,15,0.35)
[1] 0.2123387
> choose(15,5)*0.35^5*0.65^10
[1] 0.2123387
> table(rbinom(1e6,15,0.35)==5) FALSE TRUE
787498 212502
(c) What is the probability that the experiment succeeds more than 7 times? [that is, what is the probability that `r' is greater than or equal to 7?] Give a numerical value. (5 marks)
> 1-pbinom(7,15,0.35)
[1] 0.1132311
> sum(dbinom(8:15,15,0.35))
[1] 0.1132311
> table(rbinom(1e6,15,0.35)>=7) FALSE TRUE
754766 245234
(d) [harder] Use the Gaussian approximation to estimate the probability using pnorm() (5 marks)
> pnorm(7,5.25, 1.847295)
[1] 0.828265
In binomial distribution, (n=15,p=0.35,r=7)