sundaylinus:

pls. Benny put more light, this topic was introduced to us by this assignment and we are to submit it on 20th pls.

ok bro , didn't have much time , was in the lecture hall .
But go get a probability /statistics text book for details .

Now on the question , considering 500 families ( maybe in Lagos ) we're told that each of them have 5 children . could be all boys, or all girls or mixed .Using probability techniques , we want to predict / say on average the number of boys the family could consist of .

we use binomial distribution ,

p= probability of a boy in a family

q= probability of NOT a boy ( i.e a girl)

since we know that most usually , you're either a boy or a girl ( ignore special cases ) , just like tossing a coin , we either have 'heads' or 'tails' ( i.e only two possible outcomes ) , thus in that case , we have equal probabilities

we should know that p+q=1 ...........( from axioms)

p=1/2 ........(probability of success )

q= 1 -p = 1- 1/2= 1/2 ....(probability of failure )

now , we see that

p=q=1/2

from here go to my initial solution above .

follow it carefully and finish up , you don't expect me to do everything for you do you.? or you tell me where you're confused .


gracia...

sundaylinus:
pls help with this...
Out of 500 families with 5 children each, how many would you expect to have:
(a)exactly 2 boys
(b)more than 1 boy
(c)less than 2 boys
roundup ur answer tothe nearest whole number.

Lets use

E(x) = m =Np

known that 1= q+p

P(boy) =P(girl) =1/2

1/2 =p =q

N=500

n=5

by binomial distribution

P(X=r) = nCr*q^(n-r) * p^r

a) P(X=r=2) =5C2*(1/2)⁵ = 10/32

we would expect (500)(10/32) =5000/32 … finish up

b) P(r>1) =p(2) +p(3)+p(4) +p(5) = .... finish up then multiply your answer by 500

c) p(r <2) = p(0) +p(1) ..... similar as former




hope you get...

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