Questions April 2000

Questions from April 2000 paper:
1)Q.no:8,Please tell me ,Can I solve this question following way?
X has mean =3.6 & SD =2.6
P[100X>400] =P[ X>4]
=P[ X>4.5]
=P[ z>(4.5-.36)/2.6]
=P[z>.34615]
=.36461
2)-Q.no 9(i),how can solve this part ,if we take Sumation P(x=i)=1 then
alpha term =0

 

April 2000

Neetu

That's the way i understand it but there might be alternative better ways.

Q8) the qu says nothing about Normal but since n is large and mean and SD known we are allowed to use Normal dis using CLT

E(100X) = 100x 3.6 = 360
V (100X) = 10000 x 2.6*2.6 = 67,600

Hence the 100 sample follows a normal dis ( 360 , 67600/100) by CLT

P(X > 400 ) = P ( Z > (400-360)/26) and so on


Qu 9)

I got very confused with that as sum = 0

But separately say i=0 Prob should be 0< 0.2 -2a < 1

Do it for all of them and look at the possible values of a ( as if subset)

It looks very weird for one mark