Essential question: Why is the Normal distribution so special?
Here's a teaser for you--if you know that 20% of the data in a normally-distributed population fall below the x-value 306 and 80% fall below the value 772, can you find the mean and standard deviation of the population???
I will be in the classroom as early as possible on Thursday, but I have a parent meeting at 7:45. Please use your other resources well. Don't forget the book! The chapter summaries are great tools. Do the practice quiz online. Dream up questions that I might ask.
Have a slice of pizza for me.
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5 comments:
On a break during class. . .
LHSKicker: Why are you expecting Q3 to be exactly 1 SD above the mean? It should be .67 SD above the mean according to what we did in class!
Everyone else, except Joe and Chris:
How many SD away from the mean must we be to have those percentages to the left of the observed values? Use the z values to evaluate distance between the two observations.
Thanks, 2005-6 statalum-you have another shot at it here.
It's just roundoff. Statalum's answer is only correct because he exploited the fact that the measures were symmetrically distributed around the mean. (Pretty clever!) Therefore, the average of the two values worked as the mean. What if it wasn't symmetric??? The systems of equations method would still work.
Also, you know that the difference between the x-values represented a difference of 2*(.845)standard deviations. Then the length of a standard deviation must be (difference of x values)/(difference of z-values).
Good work, ladies. You are my favorite students. Get some quality sleep.
Sunday, 2-4. Thank you for remaining connected.
C-U
Yes.
p=last Sunday
q=this Sunday
pVq = true
In fact, in this case, p intersect q = true.
I stand by my answer.
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