Color blindness
The
frequency of color blindness (dyschromatopsia) in the
Caucasian American male population is about 8%.
Caucasian American male population is about 8%.
We
take a random sample of size 125 from this population. What is the probability
that six individuals or fewer in the sample are color blind?
pSampling distribution of the count X: B (n = 125, p = 0.08) à np =
10
P(X ≤ 6) = BINOMDIST(6, 125, .08, 1) = 0.1198 or about 12%
P(X ≤ 6) = BINOMDIST(6, 125, .08, 1) = 0.1198 or about 12%
pNormal approximation for the count X: N (np = 10, √np(1 − p) = 3.033)
P(X ≤ 6) = NORMDIST(6, 10, 3.033, 1) = 0.0936 or 9%
Or z = (x - µ)/σ = (6 − 10)/3.033 = -1.32 à P(X ≤ 6) = 0.0934 from Table A
P(X ≤ 6) = NORMDIST(6, 10, 3.033, 1) = 0.0936 or 9%
Or z = (x - µ)/σ = (6 − 10)/3.033 = -1.32 à P(X ≤ 6) = 0.0934 from Table A
The normal approximation is reasonable, though not perfect. Here p = 0.08 is not close
to 0.5 when the normal approximation is at its best.
A sample size of 125 is the smallest sample size that can allow use of
the normal approximation
(np = 10 and n(1 − p) = 115).
