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xEnOnn

  • 2 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?

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  1. campbell_st
    • 2 years ago
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    find the average balance for the 40... 6400/40 = $160 this means that the average balance of the 40 selected exceeds $160

  2. DBhatta
    • 2 years ago
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    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
    • 2 years ago
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    using a z score \[z =(x - \mu)/\sigma\] \[z = (160 - 150)/35\] z = 0.2857

  4. xEnOnn
    • 2 years ago
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    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
    • 2 years ago
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    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. DBhatta
    • 2 years ago
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    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. DBhatta
    • 2 years ago
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    @campbell_st actually it should be below 50% |dw:1335172829891:dw|

  8. xEnOnn
    • 2 years ago
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    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. xEnOnn
    • 2 years ago
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    I don't know which is the right way to look at this problem.

  10. campbell_st
    • 2 years ago
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    DBhatta's graph sums it up

  11. DBhatta
    • 2 years ago
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    @xEnOnn are you using z test or normal distribution? The question asks about normal distribution.

  12. xEnOnn
    • 2 years ago
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    oh, is what I have done a z-test instead of a normal distribution?

  13. DBhatta
    • 2 years ago
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    yes my friend. It should be normal distribution \[z = x - u / \sigma \]

  14. xEnOnn
    • 2 years ago
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    But it has that 40 accounts info, isn't that n=40?

  15. DBhatta
    • 2 years ago
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    ah...let me see. it's becoming more interesting now.

  16. xEnOnn
    • 2 years ago
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    haha...okay, thanks!

  17. DBhatta
    • 2 years ago
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    oh yes, it's z-distribution.

  18. xEnOnn
    • 2 years ago
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    oh so we should use the other method? The one I written just now? But why?

  19. DBhatta
    • 2 years ago
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    Yes use the one you have written. We are assuming that population mean is equal to the sample mean.

  20. xEnOnn
    • 2 years ago
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    oh I see...Thanks for you help!! :)

  21. DBhatta
    • 2 years ago
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    Actually you helped me, I was off-track. It was you who made me realize. :)

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