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Mathematics 8 Online
OpenStudy (kitkat16):

Help!!! The number of books borrowed from a library each week follows a normal distribution. When a sample is taken for several weeks, the mean is found to be 190 and the standard deviation is 30. There is a ??? % chance that more than 250 books were borrowed in a week.

OpenStudy (kitkat16):

68% - 1 95% -2 99.7% - 3

OpenStudy (amistre64):

how many standard deviation is 250 from 190?

OpenStudy (amistre64):

190 + 30n = 250, what is n?

OpenStudy (kitkat16):

n=2

OpenStudy (amistre64):

without any other information given, how do use this?

OpenStudy (kitkat16):

so 95% mean+SD(n)=250

OpenStudy (kitkat16):

so it is 2 standard deviations

OpenStudy (kitkat16):

thats it?

OpenStudy (amistre64):

the probabilty of MORE than 250 we whould now by some approximation rule that 95% of the data is within 2sd of the mean this gives us 5% to split between the tails. 2.5% each MORE than 250 is only 1 tail .... so it only represent 2.5% of the data

OpenStudy (amistre64):

your options, im assuming that what they are, do not line up with the question being asked

OpenStudy (amistre64):

or if they are just a reminder of the rule, then we have solved it

OpenStudy (kitkat16):

oh ok i need to find the chance of more than 250

OpenStudy (amistre64):

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