ISI 2011 ME 1 Q.24,27&30

[dhruv]

Please, someone help with these questions

[dhruv]

I got the solution for 24 at,

http://discussion-forum.2150183.n2.nabble.com/isi-maths-td7239469.html

Please, someone help in others.

[Rajat]

Q. 30

E(y2) = P(Y =1) *1 + P(Y=0)*0 = P(Y=1)

So basically, the problem boils down to finding P(Y=1)
Now,
Y = 1
iff   X1 + X2 + X3 + X4 ................ Xn = 100.
The only way for this expression to be = 100 is by having exactly 100 terms out of n terms where Xi =1
for each of these 100 terms.

This again boils the problem down to having a situation where X is a binomial random variable that can take values 1,0 with probabilities (p,1-p) respectively.
and us trying to find out probability of the event where we get 100 successes out of n trials.

this probability is given by nC100 *P^100 * (1-p)*n-100

[dhruv]

Thanks, Rajat. Also, can you provide some hint on Q.27, I am weak in Statistics, so I don't know all the symbols?

[Rajat]

Hey Dhruv, will try that question.
Meanwhile, can you address my question here:

http://discussion-forum.2150183.n2.nabble.com/ISI-2010-td7354483.html#a7596354

[Rajat]

Hi Dhruv, I think for q.27 the only thing we can say is that the two brands are not the same simply by the fact that their mean and variance is different. The samples collected produce different probablity distributions. Both may be normal bell curves but their form will be different.

[Yo]

For  q 27
Basically the null hypothesis is:Brands equal as in their mean of wear and tear results is equal
Alternative hypothesis : brands not alike
Now z=(x1bar-x2bar-0)/{(sx1^2/50)+(sx2^2/50)}^1/2
 Now putting values z>1.96( as confidence of 2.5% on each side) hence we reject the null hypothesis and answer is B