Question

Binomials & normals. Let X Binomial(60, 0.22) and let Y = X/60 be the proportion of successes in the sample. What is the normal distribution that approximates the distribution of Y? Calculate (approximately) the probability P(Y < 0.25) (no need to deal with continuity corrections etc). (To obtain values of "(x), the CDF of the normal distribution, you can use Table 2 in the back of your book or just search online for "standard normal table". Or, if you have R, you can use the command pnorm.)

Answer 1

Given that an B( 60,0.22) Then the post is. P(x=x) = 60 (0.227 (0.78750 -2. criven , Y = Then the post of 4 is. ve P(Y= y) =P ( 7 = 4) = P(x=604). - GO.CO, (02226040.48960-coy Y = 0, To» Zo ..! since &NB (60,0.22) -B(0,) E(X) = np. 60x0.22 = 13.2. Variane (X) = npli-p) = 60 x0.22 X0.78 = 10.296. Then E(V)- El.) - Ġo E(X) - 13.2 = 0.22 h 60 Variance of YeV (4) V(X) 100324 = 0.00286

We have Y- E(Y) = h ~ N(0,1) Var (4) - 0.25 – 0.22 Plyt 0.25) = P(Z L 0.2 10.00286 0.25) - Plz =P(2 2 0.5680) = 0.5+0.2126 - 0.7126