document.write( "Question 878778: Mike's Resume shop specializes in creating resumes for students. A recent survey of Mike's shop revealed that a spelling error was made on 10% of the resumes last year. Calculate the probabilities that a certain number of resumes have spelling errors assuming this rate continues into this year and they prepare 3424 resumes.
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\n" ); document.write( "\n" ); document.write( "c) Exactly 342 resumes.
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Algebra.Com's Answer #530253 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi
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document.write( "Below You will find a  similar workup
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document.write( "Particulars for this problem
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document.write( "p(error) = .10,  n = 3424, mean = 342.4 and SD = sqrt(342.4*.90) = 17.55
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document.write( "Use 342.5  and 341.5 as shown below
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document.write( "Stattrek.com gives a)..498  b) .502 c) .023
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document.write( "doing the work will give You the 4 decimal points
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document.write( "I.  recommend using a TI Calculator...
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document.write( "II. Must Learn when to recognize a Binomial Distribution, when You see one.
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document.write( "Most recognizable when You see a % of a certain event given.  Generally when samples are very Large(>1000), 
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document.write( "one uses a normal approximation to the binomial.
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document.write( "Trick is: When Using the normal approximation, one uses 'endpoints' (using .5)\r
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document.write( "a)p(error) = .10,  n = 5370, mean = 537 and SD = sqrt(537*.90) = 21.98
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document.write( "\"z%5B530.5%5D+=+%28530.5-537%29%2F21.98+=+-.2957\", \"z%5B529.5%5D+=+%28529.5-537%29%2F21.98+=+-.3412\", 
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document.write( "P(x > 530) = 1 - P(z ≤ -.2957) = 1 - .3837= .6163
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document.write( "b) P(x ≤ 530) = P(z ≤ -.2957) = .3837
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document.write( "c) P(x = 530) = P(z ≤ -.2957) - P(z ≤ -.3412) =.3837 - .3665 = .0172
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document.write( "Again, while it is not to 4 decimal points.. one can verify this using stattrek.com (Binomial distributions)
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document.write( "If Using TI...Using syntax: normalcdf(smaller, larger, µ, σ).
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document.write( "normalcdf(-9999, 530.5, 537, 21.98) and normal(-9999,529.5,537, 21.98)
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document.write( "would have made short work of this
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document.write( "Note: The -9999 is used as the smaller value to be at least 5 standard deviations from the mean. \n" );
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