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:
a)at least 100 people are color blind
b)exact 100 people are color blind
Thanks.
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?
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.
The available options would therefore be to say that P(Z)=1 for Z>2.9, and P(Z)=0 for Z<-2.9.
To find the probabilities using the normal approximation, we can convert the problem into a binomial distribution. In this case, the sample can be considered a binomial distribution because each person in the sample is either color blind or not color blind.
a) To find the probability that at least 100 people are color blind, we need to calculate the cumulative probability from 100 to the maximum possible value, which is 270.
The formula for the binomial distribution is:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
P(X=k) represents the probability of getting exactly k successes.
C(n, k) represents the number of combinations of n things taken k at a time.
p represents the probability of success on a single trial.
n represents the number of trials.
In this case, p = 0.20 (probability of being color blind), n = 270 (sample size), and we are interested in finding P(X >= 100).
Using the normal approximation, we can reframe the problem in terms of a standard normal distribution:
X ~ N(np, np(1-p))
Where np = 270 * 0.20 = 54, and np(1-p) = 270 * 0.20 * (1-0.20) = 43.2
Now, we need to calculate the z-score for 100 using the formula:
z = (x - mean) / standard deviation
z = (100 - 54) / sqrt(43.2)
z = 2.61
We can then use a standard normal distribution table or calculator to find the cumulative probability P(Z >= 2.61). The cumulative probability will give us the probability that at least 100 people are color blind.
b) To find the exact probability that exactly 100 people are color blind, we can use the same formula as before:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
P(X=100) = C(270, 100) * 0.20^100 * (1-0.20)^(270-100)
Calculate the combination term C(270, 100) = 270! / (100! * (270-100)!)
Using a calculator with a combination function or a statistical software, find the combination term.
Finally, calculate the probability P(X=100).
Note that the normal approximation may not be accurate for exact probabilities since it assumes a continuous distribution. For more precise results, use the exact binomial distribution formula.