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

Question

a)at least 90 people are colour blind
b)exactly 50 people are colour blind
I already posted this question and my solution. However, I am confused how do I find the a) and b) using the normal aaproximation.
My solution is:
n=270
p=0.2
np=54
np(1-p)=54(1-0.2)=43.2
z=(90-54)/6.57=5.479
P[x>90] ?
p[x=50] ?
Please explain how to find p[x>90] and p[x=50]
Thank you in advance.

Answer 1

The proper way to calculate P(x≥90) is to sum the binomial probabilities from 90 to 270. Using the computer, it should not be a problem. Using a calculator will need a little patience.

For P(x≥90), np>10, npq>10 and with a sample of 270 people (n>>40), you can use the normal approximation.
You have already calculated μ and σ² so that should not be a hurdle.
However, you need to do the continuity correction, which means that you would calculate P(x>89.5) using the normal distribution. The difference of the upper tail is about 50% with or without the correction, but both numbers are very small (around 10^(-8)).

For P(x=90), the binomial distribution would be required.
P(x,n,p)
=C(n,x)*p^x*(1-p)^(n-x)

Answer 2

Thanks. However, I've found z-score value and it's greater than 3.
z=(89.5-54)/6.57=5.4
If I can't find the value in the Z table, how I'm supposed to find the area value? I know I'm supposed to subtract the area value from 1 to find the probability for x>=90.
Please explain. Thank you very much.

Answer 3

Quite true. If you do not have access to software nor calculator, this might be a problem using paper tables.
You can try the following:
1. following link gives probabilities for z=0,1,2...50
but does not give fractional z values.
2. if possible, find a calculator that gives the normal distribution, or software. You can download (free) the R software at the following link:
https://www.r-project.org/
It is a serious and respected software, and is not too early to start using it if you are anywhere close to seeing statistics in your future career.
3. I created an abridged table for Z=3.0 to 7.9 for your reference.
The horizontal line below "3" is for probabilities for z=3.0, 3.1, 3.2...3.9 etc.
4. Abandon the normal approximation, and calculate P(90)+P(91)...P(270) using the binomial distribution.
Consolation: You can stop adding as soon as P(x) is much smaller than P(90).
In fact, if you sum from P(90) to P(100), you will get 4 significant digits of accuracy.

Answer 4

Here's the table of P(Z') for Z=3.0 to 7.9 in steps of 0.1
3
1.3498E-3 9.676E-4 6.8713E-4 4.8342E-4 3.3692E-4 2.3262E-4 1.591E-4 1.0779E-4 7.2348E-5 4.8096E-5
4
3.1671E-5 2.0657E-5 1.3345E-5 8.5399E-6 5.4125E-6 3.3976E-6 2.1124E-6 1.3008E-6 7.9332E-7 4.7918E-7
5
2.8665E-7 1.6982E-7 9.9644E-8 5.7901E-8 3.332E-8 1.8989E-8 1.0717E-8 5.9903E-9 3.3157E-9 1.8175E-9
6
9.8658E-10 5.3034E-10 2.8231E-10 1.4882E-10 7.7688E-11 4.0159E-11 2.0557E-11 1.042E-11 5.2309E-12 2.6001E-12
7
1.2798E-12 6.2372E-13 3.0109E-13 1.4388E-13 6.8167E-14 3.1974E-14 1.4876E-14 6.8833E-15 3.1086E-15 1.3322E-15

Note: P(Z)=1-P(Z')

Answer 5

Which sentence correctly describes a data set that follows a normal distribution with a standard deviation of 4 and a mean of 14?

68% of the data points lie between 10 and 14.

68% of the data points lie between 8 and 12.

68% of the data points lie between 10 and 18.

68% of the data points lie between 10 and 16