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**cynthia258****Guest**

A statistician conducted a research and the results showed that 65% of blue-collar workers agree that 8 working hours per day is acceptable. A random sample of n blue-collar workers was selected and let X be a random variable defined as the number of blue-collar workers agree the view.

(a)What is the minimum value of n such that X can be approximated by the normal random variable? Using this value of n,find the mean and standard deviation of the sampling distribution of the sample proportion.

(b) Based on the results of (a), find the probability that less than half of the blue-collar workers agree the view.

(c) A statistician conducted the same research for a random sample of 200 white-collar workers. It was found that 150 of them agree the view. Find the 97.5% confidence interval for the population proportion of white-collar workers who agree that 8 working hours per day is acceptable.

Thank you.:)

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,796

Hi cynthia258;

Welcome to the forum!

a) They usually say n = 30 or more. This is only a very rough estimate and their are precise bounds but n = 30 is what we will use.

mean = .65 x 30.

The standard deviation is

b) P(15 or less agree) = .04248

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**cynthia258****Guest**

excuse me, how can you know p(x<15) in part b? i don't understand why the number is 15

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,796

Hi;

n = 30 is what I picked.

This guy explains it well.

http://www.jedcampbell.com/?p=262

Also it is the magic number for picking from any distribution.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**cynthia258****Guest**

p(x<15)=p(z< 15-19.5 / 2.612 )=p(z<-1.7225)=0.5-0.4573=0.0427

why isn't my ans same with yours? is my calculation anything wrong?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,796

Hi;

For one thing (15 - 19.5 ) / 2.612 ≈ -1.7228

The rest could be just round off. Please recompute with the correct z score I gave above.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**cynthia258****Guest**

In part c,

the confidence interval is 97.5%. Then, which t-table should I look at? The 95% one or the 98% one? Sorry for disturbing and thank you again.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,796

A t - table is the students distribution. That is for small sample sizes.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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