Share
Explore BrainMass

The Distribution and Probability of Xbar

Let Xbar be the mean of a random sample of Size 5 from a normal distribution with &#956; = 0 and &#963;^2 = 125. Determine c so that Pr(Xbar < c)=0.90.

Please see the attached file for the fully formatted problems.

Attachments

Solution This solution is FREE courtesy of BrainMass!

Solution. As we know that Xbar will follows a normal distribution with mean mu=0 and the standard deviation sigma/sqrt(n)=sqrt(125)/sqrt(5)=5 (Note: sigma^2=125 and n=5 in this question.),
P(Xbar<c)=0.90
means that
P((Xbar-0)/5<c/5)=0.90

Note that Z=(Xbar-0)/5~N(0,1). We have
P(Z<c/5)=0.90
So,
c/5=1.282
i.e.,
c=6.41

So, the value of c is 6.41.