Consider a binomial experiment with n = 10 and p = 0.10. (a) Compute f(0). If required, round your answer to four decimal places. (b) Compute f(2). If required, round your answer to four decimal places. (c) Compute P(x ≤ 2). If required, round your answer to four decimal places.

In a certain state the proportion of families living below the poverty level is 21 %. For samples of n = 750 randomly selected families, what is the probability that the sample proportion of families that live below the poverty level is within ±0.03 from the population proportion? z=  (p-π)/√(π(1-π)/n) I got an answer of 0.9566. I am not sure if it's right. Can the problem be worked out so I can better understand it?

Consider a binomial experiment with n = 14 and p = 0.5. a. Compute f(0) (to 4 decimals). f(0) = b. Compute f(3) (to 4 decimals). f(3) = %3D c. Compute P(r 4) = e. Compute E(x) (to 1 decimal). E(r) = f. Compute Var(z) and o. Var(x) = (to 2 decimals) (to 2 decimals)

Consider a binomial experiment with n = 14 and p = 0.3. a. Compute f(0) (to 4 decimals). f(0) = b. Compute f(5) (to 4 decimals). f(5) = c. Compute P(x 2) (to 4 decimals). P(x > 2) = e. Compute E(x) (to 1 decimal). E(x) = f. Compute Var(x) and o. Var(x) = (to 2 decimals) O = (to 2 decimals)