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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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campbell_stBest ResponseYou've already chosen the best response.1
find the average balance for the 40... 6400/40 = $160 this means that the average balance of the 40 selected exceeds $160
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
X~N(150,35) X= Total balance / 40 X= 6400/40 X = 160 P(X>160) = ? =P(Z>(1600150)/35) =P(Z>0.285) = 1 P(z<0.285) =1  0.6122 = 0.3878
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
using a z score \[z =(x  \mu)/\sigma\] \[z = (160  150)/35\] z = 0.2857
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.1
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...
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
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
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
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.
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
@campbell_st actually it should be below 50% dw:1335172829891:dw
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.1
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: \[\frac { 160150 }{ \frac { 35 }{ \sqrt { 40 } } } =1.807\]
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.1
I don't know which is the right way to look at this problem.
 one year ago

campbell_stBest ResponseYou've already chosen the best response.1
DBhatta's graph sums it up
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
@xEnOnn are you using z test or normal distribution? The question asks about normal distribution.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.1
oh, is what I have done a ztest instead of a normal distribution?
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
yes my friend. It should be normal distribution \[z = x  u / \sigma \]
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.1
But it has that 40 accounts info, isn't that n=40?
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
ah...let me see. it's becoming more interesting now.
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
oh yes, it's zdistribution.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.1
oh so we should use the other method? The one I written just now? But why?
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
Yes use the one you have written. We are assuming that population mean is equal to the sample mean.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.1
oh I see...Thanks for you help!! :)
 one year ago

DBhattaBest ResponseYou've already chosen the best response.2
Actually you helped me, I was offtrack. It was you who made me realize. :)
 one year ago
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