6. 15 points) We are rolling a fair die, and keeping a running sum (that is, the

Question

6. 15 points) We are rolling a fair die, and keeping a running sum (that is, the sum of all of the results so far). Let X be the number of rolls it takes for this sum to be 100. The pmf for X is complicated; in this question, we will esti ate it through simulation using MATLAB lm. Note that X S 100. Here is a summary of what we will do Allocate an array of length 100 called px that will hold our estimates of px(k) for k 31,..., 100. (Note that the first 16 entries will always remain 0.) Set the number of experiments Q. Each experiment consists of rolling the die until the sum is 100 Loop over Q iterations. At each iteration. we set the roll count x 0 and the running sum sum 0, then generate random die rolls, incrementing sum by the result and x by one after each, until sum 100. This inner loop is easily accomplished with a while loop. A single die roll can be simulated using2 ceil (6*rand(1)) Increment the corresponding entry of px.

Explanation / Answer

Code is below:

% Experimental pmf of number of rolls it takes to get a sum >= 100

%% this code is simple but slowtic

Q = 100000;

pX = zeros(100,1);

for qq = 1:Q

runningSum = 0;

x = 0;

while (runningSum < 100)

runningSum = runningSum + ceil(6*rand(1));

x = x + 1;

end

pX(x) = pX(x) + 1;

end

pX = pX/Q;

toc

%% this code is much faster, but harder to understandtic

Q = 100000;

pX = zeros(100,1);

for qq = 1:Q

cP = cumsum(ceil(6*rand(100,1)));

x = find(cP >= 100, 1);

pX(x) = pX(x) + 1;

end

pX = pX/Q;

toc stem(pX,’LineWidth’,2);

set(gca,’FontSize’,20);

xlabel(’x’,’FontSize’,20); ylabel(’p_X(x)’,’FontSize’,20);

EXECUTE THE CODE AND YOU WILL GET THE PLOT

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