I have a problem; Let X1 + X2 + ... + X20 be independent random variable, alluni

Question

I have a problem;

Let X1 + X2 + ... + X20 be independent random variable, alluniformly distributed on (0,1)

A) Find the mean (mu) and variance (sigma^2) of the Xj's.
B) Use the central limit theorem, and a normal table, toapproximate the probability that X1 + ... X20 remains between 8.5and 11.7.

For this problem on A) I took both the mu = 1/2(1+0) = 1/2 sincethat's the mu for uniform variables (mu=1/2(b+a), likewise I tooksigma^2=var=1/12*(b-a)^2=1/12*(1-0)^2=1/12.

For this problem on B) I need to calculate:
Y = X1 + X2 ... + X20;

P{8.5 <= Y <= 11.7 }
mu(Y) = 20*mu = 20*(1/2)=10
sigma^2(Y) = 20*sigma^2 = 20/12

So using our Z-Index transform we have z-index = (x - mu)/sigma soto figure out on a standard normal distribution our ranges wouldbe:

theta1 = (8.5-10)/sqrt(20/12) = -1.1619
theta2 = (11.7-10)/sqrt(20/12) = 1.3168

Where our probability on a normal distribution (Mu=0,Sigma^2=1)

P{theta1 <= z <= theta2} OR

P{-1.1619 <= z <= 1.3168}

Based on the tables (my table just calculates the probability fromthe mean center to the inputted value):

NormalTable(theta1) = 0.3770
NormalTable(theta2) = 0.4049

So our probability would be P = 0.3770 + 0.4049 = 0.7819 =78.19%

Does this sound correct?

Explanation / Answer

For part A, since the variables are all iid, and you have 20variables, you just have to multiply the mean for one of them (1/2)by 20....the mean of X1 + X2....+X20 = 10. The variance isthe same because of independence (add the variances). Once you have those values, part b should be muchsimpler.

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