# statistics help

• Nov 30th 2012, 03:04 PM
dumbledore
statistics help
CREA puts the average price at which a house was sold in Canada in 2010 at $339,100. Let us assume that the standard deviation of the house prices is$134,216. Suppose you had done a survey of a random selection of 36 house sale prices in 2010 and obtained an average house price of $368,533 /SOLVED • Nov 30th 2012, 03:50 PM abender Re: statistics help Quote: Originally Posted by dumbledore CREA puts the average price at which a house was sold in Canada in 2010 at$339,100. Let us assume that the standard deviation of the house prices is $134,216. Suppose you had done a survey of a random selection of 36 house sale prices in 2010 and obtained an average house price of$368,533
What is the chance that such a survey would have resulted in an average price this high or higher.

$\displaystyle P\left(X>368533\right) = 1 - P\left(X\leq368533\right)$

How do you switch from $\displaystyle P\left(X\leq \text{whatever}\right)$ to $\displaystyle P\left(Z\leq\text{whatever}\right)$?

Hint: $\displaystyle Z=\frac{X-\mu}{\sigma/\sqrt{n}}$
• Nov 30th 2012, 03:59 PM
dumbledore
Re: statistics help
I get z=1.32. Is that the answer?
• Nov 30th 2012, 04:29 PM
abender
Re: statistics help
Quote:

Originally Posted by dumbledore
I get z=1.32. Is that the answer?

It is not the "answer" to the problem, but it is the value in the Z-table that you will look up.

The answer is $\displaystyle 1 - P(Z\leq1.32) \approx 1 - 0.9066 = 0.0934$.
• Nov 30th 2012, 04:33 PM
abender
Re: statistics help
Also, for these types of problems, ask yourself if the answer makes sense. Use 68-95.5-99.7 as a guideline.