Thread: Normal Approximation to the Binomial Distribution

Hey! I really need some help, I have no clue how to do these:

1) A manufacturer of pencils has 60 dozen pencils randomly chosen from each day's production and checked for defects. A defect rate of 10% is considered acceptable. Assuming that 10% of all the manufacturer's pencils are actually defective, what is the probability of finding 80 or more defective pencils in this sample?

2) A store manager believes that 42% of her customers are repeat business (have visited the store within the last two weeks). Assuming that she is correct, what is the probability that out of the next 500 customers, between 200 and 250 customers are repeat business.

Any help is much appreciated. Thanks a ton in advance!

For normal distribution approximation you can find z-scores to be

$\displaystyle Z=\frac{x-np}{\sqrt{npq}}$

In the first problem

$\displaystyle n=720, p = \frac{1}{10} , q = 1-p$

$\displaystyle Z=\frac{80-720\times \frac{1}{10}}{\sqrt{720\times \frac{1}{10}\times \frac{9}{10}}}
$

$\displaystyle Z=\frac{8}{8.05}$

$\displaystyle Z=0.99$

Using normal dist table find $\displaystyle P(Z>0.99)$

In the second problem find

$\displaystyle P(X<250)-P(X<200)$

You must change both of these into z-scores as above with parameters

$\displaystyle n = 500, p= \frac{42}{100}, q = 1-p$