normal distribution part a and b

[maths_arghh234]

this is the big question and i got both parts wrong

7. samples of blood was taking were taken from 250 children in a region in India. of these children, 4 had blood type A2B , write down an estimate of p , the proportion of all children having the blood type A2B.
consider a group of n children from the region and let X be the number of blood A2B.
assuming that X os distributed B(n,p) and that p has the value estimated above calculate to 3 d.p the propability that the number of children with the blood type A2B in a group of 6 children from the region will be
a . 0
b . more than 1
use a poisson distribution approximation to calculate , to 4 d.p the propbability that in a group of of 800 children from this region, there will be fewer than 3 children of blood type A2B

i have done a actually
(just realized)
a. p(x=0) =0.908

 

[maths_arghh234]

is anyone going to help me?

 

[galactus]

7. samples of blood was taking were taken from 250 children in a region in India. of these children, 4 had blood type A2B , write down an estimate of p , the proportion of all children having the blood type A2B.

4/250=.016

the propability that the number of children with the blood type A2B in a group of 6 children from the region will be

a . 0

Correct. It is about .908

b . more than 1

You can use the binomial as in part a, except find the cases for k=0 to 1, then subtract from 1:

\(\displaystyle 1-\sum_{0}^{1}\binom{6}{k}(\frac{2}{125})^{k}(\frac{123}{125})^{6-k}\)

use a poisson distribution approximation to calculate , to 4 d.p, the propbability that in a group of of 800 children from this region, there will be fewer than 3 children of blood type A2B

Since the probability from the first part is .016=2/125, then \(\displaystyle {\lambda}=800(.016)=12.8\)

So, we want fewer than 3. Add up the cases x=0 to 2

\(\displaystyle \sum_{k=0}^{2}\frac{{\lambda}^{x}\cdot e^{-\lambda}}{x!}\)