Normal Approximation to the Binomial Distribution

**Sculthorp** · May 31st 2009, 01:07 PM

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!

**pickslides** · May 31st 2009, 03:24 PM

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

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

In the first problem

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

$Z=\frac{80-720\times \frac{1}{10}}{\sqrt{720\times \frac{1}{10}\times \frac{9}{10}}}<br />$

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

$Z=0.99$

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

**pickslides** · May 31st 2009, 03:28 PM

In the second problem find

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

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

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

Thread: Normal Approximation to the Binomial Distribution

LinkBack

Thread Tools

Display

Normal Approximation to the Binomial Distribution

Similar Math Help Forum Discussions

normal approximation to the binomial

To make a normal approximation for a binomial distribution

Normal approximation to binomial distribution.

Question regarding the normal approximation to the binomial probablility distribution

Normal approximation to the binomial distribution

Search Tags