SOLUTION: Suppose that you work for a newly restructured automotive company with nearly 100,000 employees. You are in charge of purchasing engines from an overseas supplier. Company policy i

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Question 1136800: Suppose that you work for a newly restructured automotive company with nearly 100,000 employees. You are in charge of purchasing engines from an overseas supplier. Company policy is that you purchase several hundred engines each month to be placed into cars on the assembly line.
Your overseas supplier of engines guarantees that 10% of the new engines shipped to you will have minor oil leaks that will require a slight modification before assembly. To check out the most recent monthly shipment of the engines, you randomly select and test 200 of these engines. What is the average number of leaky engines you would expect to get?
Answer I got from this was :20
Second part of the question it asks is For the same overseas supplier as the previous question, what is the standard deviation of the number of leaky engines in a random sample of 200 shipped engines? What do I do for this?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
mean = n * p = 200 * .2 = 20

standard deviation = sqrt( n * p * q ) = sqrt( 200 * .2 * .8 ) = 5.656854249.

i believe that's the answer you're looking for.

here's a reference.

https://stattrek.com/probability-distributions/binomial.aspx

in your binomial distribution:

n = 200
p = .1
q =.9
mean = 200 * .10 = 20
standard deviation is sqrt (200 * .2 * .8) = 5.656854 rounded to 3 decimal places.

the variance is equal to n * p * q.
the standard deviation is the square root of the variance which makes it equal to sqrt (n * p * q).

this forms a normal distribution where the mean is 20 and the standard deviation is 5.656854.

about 68.3% of the defects will be centered within 1 standard deviation from the mean.

$$$

about 95.4% of the defects will be centered within 2 standard deviations from the mean.

$$$

about 99.7% of the defects will be centered within 3 standard deviations from the mean.

$$$

the normal distribution calculator i used can be found at http://davidmlane.com/hyperstat/z_table.html