Question

Consider a box which contains some tickets. Suppose a fraction, p, of the tickets have a "1" written on them; the remaining have a "0". a. Let X denote the number on a ticket which is randomly drawn from the box. Fill in the first blank in the following sentence with the expected value of X and the second blank with the standard deviation of X. We expect X to be P give or take sqrt(p(1-p)) Write a few sentences to explain how you calculated the numbers. You could include a table showing the probability distribution of X and then show how to calculate E(X) and SD(X) from the entries in the table b. Let X denote the fraction of 1's in 100 draws made with replacement from the box. That is, + X 100 X = X1 + X2 + 100 where X¡ denotes the number on the ith draw. Fill in the first blank in the following sentence with the expected value of X and the second blank with the standard deviation of X. (To write the standard deviation symbolically, use the dollar sign like this:Vablc for instance) We expect X to be give or take Write a few sentences to explain how you calculated the numbers. You could include rules you used to arrive at your answers as well.

Answer 1

a)

The experiment explained is a Bernoulli trial.

Let there be n tickets with p of them having 1 and n-p having 0

The expectation of X is calculated as:

p*1+(n-p)*0 = p

The expectation of X^2 is calculated as:

p*1+(n-p)*0 = p

The variance of X is calculated as:

E(X^2) - E(X)^2

= p - p^2

= p*(1-p)

The standard deviation of X is sqrt [ p*(1-p) ]

So, the expectation of X is p ,.

Standard deviation of X is sqrt [ p*(1-p) ]

We expect X to be p give or take sqrt [ p*(1-p) ]

b.

The experiment is similar as above

Here we have 100 tickets , that is n = 100

x̅ is the fraction of 1's

x̅ = [ X1+X2+...+X100 ] / 100

Since Xi's are independent, expectation of x̅ is calculated as:

[ E(X1)+E(X2)+...+E(X100) ] / 100

= [ p + p + ... + p ] / 100

= 100* p /100

=p

The variance of x̅ is calculated as:

Var(x̅) = [ Var(X1)+Var(X2)+...+Var(X100) ]/100^2

= 100*p*(1-p)/100^2

=p*(1-p)/100

The standard deviation of x̅ is sqrt [ p*(1-p)/100 ]

So, the expectation of x̅ is p ,.

Standard deviation of x̅ is sqrt [ p*(1-p)/100 ]

We expect x̅ to be p give or take sqrt [ p*(1-p)/100 ]