Probability problem

• Sep 30th 2009, 07:22 PM
Almondzqueen
Probability problem
Here's a problem I'm stuck with:
A survey for brand recognition is done and it is determined that 70% of consumers have heard of Mike's Mechanic Shop. A survey of 900 randomly selected cosumers is to be conducted. For such groups of 900, would it be unusual to get 527 consumers who recognized this shop? So I have to show all the statistics.....

So I'm not clear whether I divide 527/900? Or do I multiply (900)*(.70)?
And why would it be unusual? Do I have to draw a bell shaped curve as well (what would it look like)?
I need this explained step by step.
Thank you.(Thinking)
• Sep 30th 2009, 08:18 PM
pickslides
So there was an initial survey that gave the figure of 70% and now there is another survey that will have 900 participants?

$\displaystyle \frac{527}{900} \approx 59\%$

you mentioned a bell shaped curve, are you given a standard deviation?
• Oct 1st 2009, 02:14 PM
awkward
Quote:

Originally Posted by Almondzqueen
Here's a problem I'm stuck with:
A survey for brand recognition is done and it is determined that 70% of consumers have heard of Mike's Mechanic Shop. A survey of 900 randomly selected cosumers is to be conducted. For such groups of 900, would it be unusual to get 527 consumers who recognized this shop? So I have to show all the statistics.....

So I'm not clear whether I divide 527/900? Or do I multiply (900)*(.70)?
And why would it be unusual? Do I have to draw a bell shaped curve as well (what would it look like)?
I need this explained step by step.
Thank you.(Thinking)

Assuming the figure of 70% is correct, the number of customers in the survey who have heard of the shop should have a Binomial distribution with p = 0.70 and n = 900. The mean of the distribution is np = 630. So I would interpret the question as asking what is the probability that 527 customers or fewer will recognize the shop. I.e., if X has a Binomial distribution with p = 0.70 and n = 900, what is the probability that $\displaystyle X \leq 630$? Oops-- that should be 527.[/edit]

You certainly wouldn't want to calculate this probability by hand, but you might have a calculator that will do it for you. Or you could use a Normal approximation to the Binomial-- that would be my choice.
• Oct 2nd 2009, 08:23 PM
Almondzqueen
Quote:

Originally Posted by pickslides
So there was an initial survey that gave the figure of 70% and now there is another survey that will have 900 participants?

$\displaystyle \frac{527}{900} \approx 59\%$

you mentioned a bell shaped curve, are you given a standard deviation?

I actually remember her mentioning standard deviation but I can't find my notes that go with it! I do clearly remember her saying to use binomial distribution though. =/
• Oct 3rd 2009, 09:58 AM
awkward
Please note that I wrote down a wrong number in my original post, which I went back and corrected.

If you don't know the formula for the standard deviation of a binomial distribution, it is $\displaystyle \sqrt{np(1-p)}$.

See
Binomial distribution - Wikipedia, the free encyclopedia