Statistics Help! Urgent!

Question

Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 130 people from the U.S. is chosen, approximate the probability that at least 20 are left-handed. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.

Answer 1

Here, np is 130(0.16) or 20.8 and nq is 130(1 − 0.16) or 109.2; both results exceed 5 so the normal approximation can be used.

Take μ and σ for the Binomial Distribution as np or 20.8 and √[npq] or √[130×0.16×0.84] equal to

√17.472.

The problem calls for P(x ≥ 20) which is corrected for continuity by P[x ≥ (20 − 0.5)] or P(x ≥ 19.5).

Take z_19.5 as (19.5 − 20.8) ÷ √17.472 or -0.3110083447.

-0.3110083447 corresponds to an area (or probability) under the Standard Normal Curve of -0.1221028611.

P(x ≥ 19.5) is then written as (1 − 0.1221028611) or 0.8778971389 equivalent to 0.878.