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Mathematics 21 Online
OpenStudy (anonymous):

betty's bite-size candies are packaged in bags. The number of candies per ba is normally distributed, with a mean of 50 candies and a standard deviation of 3. At a quality control checkpoint, a sample of bags is checked, and 4 bags contain fewer than 47 candies. How many bags were probably taken as samples?

OpenStudy (amistre64):

hmmm, im wanting to say this involves a sqrt(n)

OpenStudy (amistre64):

\[z=\frac{x-\bar x}{sd}\sqrt{n}\] \[\frac{z(sd)}{x-\bar x}=\sqrt{n}\] \[(\frac{z(sd)}{x-\bar x})^2=n\] but im not too sure if thats a good thought yet or not

OpenStudy (amistre64):

there are 47 is wiithin 1 sd from the mean which is about what we would expect 16% of the sample to have since there are 4 bags .... then 100 is to 16 as 25 is to 4

OpenStudy (amistre64):

might be right :) any ideas?

OpenStudy (amistre64):

50 - 3 = 47, is 1 sd from the mean. 1 sd from the mean as about 34% on each side 50 - 34 = 16, so 16 out of 100 are expected to have less than a 47 count. 4 is 16% of what number?

OpenStudy (amistre64):

\[\frac{16}{100}=\frac{4}{n}\] \[\frac{4}{100}=\frac{1}{n}\] \[\frac{100}{4}=n=25\]

OpenStudy (anonymous):

can you help me in another question

OpenStudy (anonymous):

please

OpenStudy (amistre64):

dunno, my smarts are fairly limited to this question and something about marginal tax flows in algeria

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