It is known that approximately 20% of the population is colour blind. In a sample of 270 people, use the normal approximation to find the probability that:

Question

a)at least 100 people are color blind
b)exact 100 people are color blind
Thanks.

Answer 1

n=270 (>40)
p=0.2
q=1-0.2=0.8
np=270*0.2=54 (>10)
nq=270*0.8=216 (>10)

All prequisites for normal approximation are satisfied.

μ=np=54
s=std dev=√(npq)=sqrt(43.2)

a)
P(X≥100)
apply continuity adjustment
Z=(99.5-μ)/s=6.92
P(Z≥6.92)~1-2.26*10^(-12)
=1 approximately.

b)P(X=100)
apply continuity adjustment
Z2=(100.5-μ)/s=7.0747...
Z1=(99.5-μ)/s=6.9226...
P(X=100)
=P(Z2)-P(Z1)
=(1-P(Z2'))-(1-P(Z1')
=P(Z1')-P(Z2')
=(2.217072-0.748594)*10^-12
=1.468*10^(-12)
~0

Note:
It is unfortunate that most normal probability tables (and calculators) do not go beyond Z>6, which means that special software is required. It would be interesting to know
1. is the probability for this problem 0.2 and not 0.4 or any other value?
2. if the probability is 0.2, perhaps your teacher could explain to you how the normal approximation will be evaluated. Are 1 and 0 acceptable answers?

Answer 2

Thank you so much. You have been really helpful. The teacher gave us the normal probability table that ends at 2.9 So it was impossible for me to solve without the proper software or the propability table.

Answer 3

The available options would therefore be to say that P(Z)=1 for Z>2.9, and P(Z)=0 for Z<-2.9.