HW8 Solution

Note: only one problem was *'d (my mistake) and so this homework is out of a total of 6 points.

Chapter 18, #20

I can't show you  a picture, but I'll try to derive it.  The idea here is that even though the distribution of home prices in the population is skewed right, and even though the histogram of the prices of a random sample of homes would be right-skewed, the normal probability model is a good model for the probabilities of the average of the sample.

Keep in mind that averages are random numbers as long as they're based on random samples.  If the observations are independent of each other, then the probability distribution of an average of n numbers will follow the normal model, with the mean the same as the population mean (in this case $140K) and the sd is the sd of the population divided by the square root of the sample size:    60000/sqrt(100) = $6K.

So if everyone in this class were to take a random sample of 100 homes and find their average price, 68% of us would get an average between 140K - 6K and 140K + 6K  (134000 to146000).  95% of us would get an average between 140K - 2*6K and 140K + 2*6K  or (128000 to to 152000).  And almost all of us, about 99.7%, would have an average between
140K - 3*6K to 140K + 3*6K  (122000 to 158000).

Note that it is NOT true that 68% of the houses are between 134000 and 146000.  We don't know what range covers 68% of the houses because we don't know the distribution of houses (we only know it is right skewed).  But we do know the distribution of the average of a random sample of houses.