SOLUTION: 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 t

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Question 878776: 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 8337 resumes.
For full marks your answer should be accurate to at least four decimal places.
a) More than 834 resumes.


b) At least 834 resumes.


c) Exactly 834 resumes.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

Below You will find a  similar workup

Particulars for this problem

p(error) = .10,  n = 8337, mean = 833.7 and SD = sqrt(833.7*.90) = 27.392

Use 834.5  and 833.5 as shown below

Stattrek.com gives a).448  b) .512  c) .015

doing the work will give You the 4 decimal points

...........................................................................

I.  recommend using a TI Calculator...

II. Must Learn when to recognize a Binomial Distribution, when You see one.

Most recognizable when You see a % of a certain event given.  Generally when samples are very Large(>1000), 

one uses a normal approximation to the binomial.

Trick is: When Using the normal approximation, one uses 'endpoints' (using .5)



a)p(error) = .10,  n = 5370, mean = 537 and SD = sqrt(537*.90) = 21.98

z%5B530.5%5D+=+%28530.5-537%29%2F21.98+=+-.2957, z%5B529.5%5D+=+%28529.5-537%29%2F21.98+=+-.3412, 

P(x > 530) = 1 - P(z ≤ -.2957) = 1 - .3837= .6163

b) P(x ≤ 530) = P(z ≤ -.2957) = .3837

c) P(x = 530) = P(z ≤ -.2957) - P(z ≤ -.3412) =.3837 - .3665 = .0172

Again, while it is not to 4 decimal points.. one can verify this using stattrek.com (Binomial distributions)

If Using TI...Using syntax: normalcdf(smaller, larger, µ, σ).

normalcdf(-9999, 530.5, 537, 21.98) and normal(-9999,529.5,537, 21.98)

would have made short work of this

Note: The -9999 is used as the smaller value to be at least 5 standard deviations from the mean.