5 Suppose that 10SolutionA Yes this can be considered an alm

5. Suppose that 10

A. Yes, this can be considered an almost binomial situation because the population is 20 times larger than the sample.

B. P(4 <= x<= 16) = binomcdf ( 100 , .1 , 16 ) - binomcdf ( 100 , .1 , 13 ) = 0.9716

C) The answer that used the continuity correction is slightly higher , because the continuity correction adds .5 at the upper and the lower bound.

np = (100) (.1) = 10 >= 10

n(p-1) = (100) (.9) = 90

z = 4 -10 / sqrt[(100)(.1)(.9)]

= -6 / 3 = -2

z = 16 - 10 / sqrt[(100)(.1)(.9)

= 6 / 3 = 2

P (-2 < z < 2 ) = normalcdf(-2,2) =0.9545

z = 3.5 - 10 / sqrt[(100)(0.1)(.9) = - 6.5 / 3 = -2.1667

z = 16.5 - 10 / sqrt[(100)(.1)(.9) = 6.5 / 3 = 2.1667

P(-2.1667 < z < 2.1667 ) = normalcdf( - 2.667, 2.667 ) =0.9697