Question

A survey indicated that 72% of blah if given a choice would prefer to start their own business. A random sample of 600 people is obtained. What is the probability that no more than 70% would prefer to start their own business?

Answer 1

Anyone? :(

Answer 2

Hi there

Answer 3

I can help you with this if you like

Answer 4

Actually I think I have this one, I found something in my book.

Answer 5

Thanks though

Answer 6

Just curious, how did you solve it?

Answer 7

Well I'm in the process of solving it... turning .70 into a z-score by subtracting mean and dividing by standard deviation.

Answer 8

I think that doesn't work in this case because you don't know the standard deviation of the population

Answer 9

..true.

Answer 10

This can be modelled exactly using the binomial distribution, which if you have a fancy calculator you can compute

Answer 11

Do you know the binomial distribution?

Answer 12

The... uhh... What is that again?

Answer 13

binomial distribution

Answer 14

I don't know that.

Answer 15

do you know the answers? i want to make sure my answer is right before i go on a wild goose chase teaching you

Answer 16

I can look it up in the back of the book.

Answer 17

is it 0.147989?

Answer 18

Says here P(phat < .7) = .1379

Answer 19

hmm, not sure how they got that

Answer 20

Oh well, might just skip this one.

Answer 21

ok, BANG! I totally figured out exactly what the book did, and I have an answer that's superior to theirs.

Answer 22

Superior to the math book? You are a wizard aren't you?

Answer 23

No, just a regular wizard (e.g. broomsticks)

Answer 24

Haha, very well.

Answer 25

Anyway so here's the deal. The *exact* way of computing the answer to that question is using something called the Binomial Distribution. The Binomial Distribution is another distribution, sort of like the normal distribution. The normal distribution takes "average" and "standard deviation" as parameters, but the binomial distribution is different. It takes as parameters "n" -- the number of times you try an experiment, and p -- the probability that experiment "succeeds". Then the distribution tells you what the probability is that any particular number of successes will occur.

So the *exact* answer to your question is the CDF of the binomial distribution(n=600, p=0.72) evaluated at 600*0.70. This is 0.147989.

The problem is the Binomial Distribution is a huge pain in the retriceto calculate for large values of n. I only know what it is because I used a program that can calculate it. Many people, and stodgy old textbooks, use the Normal Distribution as an *approximation* to the binomial distribution. You can approximate the Binomial using average = np, std dev = sqrt(n * p * (1-p)). When you use this approximation, and evaluate the CDF of the normal distribution with those parameters at 600*0.70, you get the book answer. So they're assuming you're going to use the normal approximation. But 0.148 is a better, more exact answer.

Answer 26

Haha, well very good. I'm flattered that you'd go to that much trouble to help me with a math problem :)

Answer 27

It's OK, I'm really here to learn math. Just exploiting hapless students like yourself to give me problems like this that make me think.

Answer 28

Well however you look at it, you've really helped, thanks a lot.

Answer 29

no problem. any other questions?

Answer 30

Uhh, yeah, small one, what's the difference between x and xbar? I'm calculating a z score and it says (x-mu)/stdev, but I'm not really sure what plain x is, I have xbar.

Answer 31

I'm not totally sure, it seems to depend on context sometimes, but I think x is a particular sample and x-bar is a sample average

Answer 32

Hmm, so should I just use the value I have for xbar?

Answer 33

probably

Answer 34

Alrighty.

Okay I've been at this for several hours, I'm gonna take a break for a bit, but I'll be back with more questions I'm sure.

Answer 35

np