Quote (Kayak @ Apr 21 2013 03:17pm)
2.
theta(sample mean)= theta/(sqt (n))= 68/(sqt(100))= 68/10= 6.8 (theta being standard deviation)
p(sample mean<10)=P(z< (10-20)/6.8)= P(z< -1.47) Where 10 is change in sample mean, 20 is mean change
Go to z table and you get 1.47 corresponding to .4292 (rounded to 3 decimals: .429)
The way you did it is for one sample. As you increase sample, the standard deviation of the sample mean goes down.
#2
Assumption: the change of scores is normally distributed
Use Z transformation. Sample 100 times. Each one has mean of 20 and sd of 68
So find that the average mean (Y) is less than 10
P(Y<10)= P(Z<(10-20)/(68/10))= P(Z<-1.67) = 1-0.9525 = 0.0475
#3
Need to see the table. Also
#5
Just the usual z transformations.
A) i) P(x>112)= P(z>(112-98)/15) = P(z>0.933) = 0.1762
etc....
ISO PAYMENT. I HOPE YOU"RE NOT A WELCHER.
This post was edited by Exx on Apr 21 2013 05:11pm