SOLUTION: I'm having a hard time understanding how to compute this question. I'm just starting my summer school class and my teacher is absolutely HORRID! It's his first time ever teaching a

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Question 878492: I'm having a hard time understanding how to compute this question. I'm just starting my summer school class and my teacher is absolutely HORRID! It's his first time ever teaching and he doesn't speak english well!
My question is how do I solve this question?
Question:
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 4057 resumes.
For full marks your answer should be accurate to at least four decimal places.
a) More than 409 resumes.
Probability:

b) At least 409 resumes.
Probability:

c) Exactly 409 resumes.
Probability:

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

Hi

I.  recommend using a TI Calculator for this summer school class ...

There are no excuses one can have for a poor grade.

Can recommend stattrek.com as an EXCELLENT reference (checking your TI results etc)

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 approxiamation, one uses 'endpoints' (using .5)

a)p(error) = .10,  n = 4057, mean = 405.7 and SD = sqrt(405.7*.90) = 19.108

P(x > 409) = 1 - P(z ≤ (++highlight_green%28409.5%29+  - 405.7)/19.1084 = 1- P(z ≤ .199) = 1 - .5789 = .4211

b) P(x ≤ 409) = P(z ≤ .199) = .5789

c) P(x = 409) = P(z ≤ .199) -  P(z ≤ (++highlight_green%28408.5%29+  - 405.7)/19.1 = .5789 - .5583= .0206

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

If Using TI...Using syntax:

normalcdf(-9999, 409.5, 405.7, 19.1) and normal(-999,408.5, 405.7, 19.1)

would have made short work of this