anonymous 4 years ago A population of normally distributed account balances have mean balance $150 and standard deviation of$35. What is the probability that the total balance for a random sample of 40 accounts will exceed $6400? • This Question is Closed 1. campbell_st find the average balance for the 40... 6400/40 =$160 this means that the average balance of the 40 selected exceeds $160 2. anonymous X~N(150,35) X= Total balance / 40 X= 6400/40 X = 160 P(X>160) = ? =P(Z>(1600-150)/35) =P(Z>0.285) = 1- P(z<0.285) =1 - 0.6122 = 0.3878 3. campbell_st using a z score $z =(x - \mu)/\sigma$ $z = (160 - 150)/35$ z = 0.2857 4. anonymous Thanks! hmm...actually, I have also calculated both the different answers by @DBhatta and @campbell_st. And that's why I am not sure what the correct answer should be. hahaha... 5. campbell_st I would have thought that if the balance is to exceed then its the probability should be between 50% and 68%... as the average is greater than the mean... and the z score is 0.2857 6. anonymous I'm pretty sure with my answer and @campbell_st also has the same answer. He has just stopped after calculating z and I've calculated until the end. 7. anonymous @campbell_st actually it should be below 50% |dw:1335172829891:dw| 8. anonymous oh no...wait, so sorry, I saw the wrong numbers. Yes, @campbell_st and$DBhatta both has the same answer. Actually, here's what I got for my other answer: $\frac { 160-150 }{ \frac { 35 }{ \sqrt { 40 } } } =1.807$

9. anonymous

I don't know which is the right way to look at this problem.

10. campbell_st

DBhatta's graph sums it up

11. anonymous

@xEnOnn are you using z test or normal distribution? The question asks about normal distribution.

12. anonymous

oh, is what I have done a z-test instead of a normal distribution?

13. anonymous

yes my friend. It should be normal distribution $z = x - u / \sigma$

14. anonymous

But it has that 40 accounts info, isn't that n=40?

15. anonymous

ah...let me see. it's becoming more interesting now.

16. anonymous

haha...okay, thanks!

17. anonymous

oh yes, it's z-distribution.

18. anonymous

oh so we should use the other method? The one I written just now? But why?

19. anonymous

Yes use the one you have written. We are assuming that population mean is equal to the sample mean.

20. anonymous

oh I see...Thanks for you help!! :)

21. anonymous

Actually you helped me, I was off-track. It was you who made me realize. :)