Sampling distributions and probabilities

BAMBER24

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According to a financial research firm, 29% of adults do not own or use a credit card. Assuming that the sample distribution of "p hat" for sample sizes of 500 is normally distributed, find the following probabilities:

a) That in a sample of 500 more than 30% do not own credit cards.

b) That in a sample of 500 between 25% and 30% do not own credit cards.

*my book only has examples where "x" is given and the probabilites they are looking for are less than. PLEASE HELP!!!! :???:
 
*my book only has examples where "x" is given and the probabilites they are looking for are less than. PLEASE HELP!!!! :???:


You can use the same methods for "less than" problems here. This is because the probability that something is at least a certain value is the same as 1 minus the probability that is it less than that value. That is, if we want to know the probability that some variable quantity \(\displaystyle X\) has at least the value \(\displaystyle k \) then

\(\displaystyle P(X \geq k) = 1 - P(X < k)\)

So if you know how to do the problem for less than, you really know how to do it for greater than also.


And if we want to know the probability that \(\displaystyle X\) is between two values \(\displaystyle k\) and \(\displaystyle j\), then we can still use the fact that we know how to compute the "less than" probabilities. That is:

\(\displaystyle P(k \leq X \leq j) = P(X \leq j) - P(X < k) \)

so it just involves two different "less than" computations.

Does that help you get started?
 
According to a financial research firm, 29% of adults do not own or use a credit card. Assuming that the sample distribution of "p hat" for sample sizes of 500 is normally distributed, find the following probabilities:

a) That in a sample of 500 more than 30% do not own credit cards.

b) That in a sample of 500 between 25% and 30% do not own credit cards.

*my book only has examples where "x" is given and the probabilites they are looking for are less than. PLEASE HELP!!!! :???:
They tell you to use a normal distribution, and they tell you the mean - but not the standard deviation. How can you use the sample size to get a standard deviation?

Do you know about the binomial distribution? That seems to be appropriate here, since people either DO of DO NOT use credit cards, p is constant (p=0.29), and there is a fixed number of trials (n=500). You must know a formula to get sigma for the binomial - then when the sample size is large enough it is safe to approximate the binomial with a normal distribution of the same mean and standard deviation.

Does that get you started? If you need more help, show us your work.
 
Every person in the sample either has and uses a credit card or does not. So that is actually a binomial distribution. The problem is that for very large n, here, 500, the binomial distribution is difficult to calculate. But we can approximate it by a normal distribution. A binomial distribution with probabilites p and 1- p and number n can be approximated by a normal distribution with mean pn and standard deviation \(\displaystyle \sqrt{np(1- p)}\).
 
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