question in april 2006

[Neetu Verma]

1)-Question 13,part(ii)b,in past question paper April 2004
Anyone tell me ,why, we take p=.8,.6,.4,.2
2)-question 5,part (ii),in paper March 2006
In alternative comment,take approximations are quite good for p=.1 But
approximation is not valid because np<5.So why takes approximations are quite good .

 

[sweetlikealemon]

1)-Question 13,part(ii)b,in past question paper April 2004
Anyone tell me ,why, we take p=.8,.6,.4,.2

For this I imagined a set of orderd observations
0 < x1 < x2 < x3 < x4 < infinity

Since we want each interval to be probability 0.2
we basically want
P(0 < X < x1) = 0.2
P(x1 < X < x2) = 0.2
P(x2 < X < x3) = 0.2
P(x3 < X < x4) = 0.2
P(x4 < X < infinity) = 0.2 or P(X > x4) = 0.2

Hence
P(X < x4) = 0.8
P(X < x3) = 0.6
P(X < x2) = 0.4
P(X < x1) = 0.2

2)-question 5,part (ii),in paper March 2006
In alternative comment,take approximations are quite good for p=.1 But
approximation is not valid because np<5.So why takes approximations are quite good .

I read somewhere that np<5 means that normal approx is not suitable, Poisson is more reliant on small p, N>= 20

Hope this helps.
Ed

 

[Neetu Verma]

1)-Question 13,part(ii)b,in past question paper April 2004
Anyone tell me ,why, we take p=.8,.6,.4,.2

For this I imagined a set of orderd observations
0 < x1 < x2 < x3 < x4 < infinity

Since we want each interval to be probability 0.2
we basically want
P(0 < X < x1) = 0.2
P(x1 < X < x2) = 0.2
P(x2 < X < x3) = 0.2
P(x3 < X < x4) = 0.2
P(x4 < X < infinity) = 0.2 or P(X > x4) = 0.2

Hence
P(X < x4) = 0.8
P(X < x3) = 0.6
P(X < x2) = 0.4
P(X < x1) = 0.2

2)-question 5,part (ii),in paper March 2006
In alternative comment,take approximations are quite good for p=.1 But
approximation is not valid because np<5.So why takes approximations are quite good .

I read somewhere that np<5 means that normal approx is not suitable, Poisson is more reliant on small p, N>= 20

Hope this helps.
Ed

Thanks for this help.