Let X_1, X_2, ..., X_n denote n random draws with replacement from a box of tick

Q: Let X_1, X_2, ..., X_n denote n random draws with replacement from a box of tick

a) E(Xi) = x*p[x] = 0 * 1-p + 1 * p = p b) V(Xi) = x2 p[x] - (xp[x])2 = 0*(1 - p) + 1 * p - p2 = pq c) E(Y) = E(x1 + x2 +........xn) = E(x1) + E(x2) + ...........E(xn) = p + p + ......... + p = np d) var (Y) = v(x1) + v(x2) + .......... + v(xn), x1, x2 .........xn are independent = pq + pq + ..........+ pq = npq

ID: 3206984 • Letter: L

Question

Let X_1, X_2, ..., X_n denote n random draws with replacement from a box of tickets where each ticket has either "0" or "1" is written on it. Let p denote the probability of picking a "1" at each random draw. In this case, we say that each random variable X_1 has a Binomial distribution based on 1 trial with probability of success p, written X_i ~ Bin(1, p). Now, consider Y = sigma_i = 1^n X_i. Then, Y simply counts the number of 1's in the list X_1, X_2, ..., X_n. In this case Y has a Binomial distribution with parameters (n, p), or Y ~ Bin(n, p). Find EX_i. Find var(X_i). Find EY using the linear formula for expected value and the answer to part (a). Find var(Y) using the linear formula for variance and the answer to part (b).

Explanation / Answer

a) E(Xi) = x*p[x]

= 0 * 1-p + 1 * p

= p

b) V(Xi) = x2 p[x] - (xp[x])2

= 0*(1 - p) + 1 * p - p2

= pq

c) E(Y) = E(x1 + x2 +........xn)

= E(x1) + E(x2) + ...........E(xn)

= p + p + ......... + p

= np

d) var (Y) = v(x1) + v(x2) + .......... + v(xn), x1, x2 .........xn are independent

= pq + pq + ..........+ pq

= npq

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Let X_1, X_2, ..., X_n denote n random draws with replacement from a box of tick

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