## Math 151 , Sp. '12, Wed., Feb. 15, Day 11 .After class. Hit reload..

HW :  Ch. 3:  Use Normal Density Applet curve to check concepts and computation. .Rest of chapter. //Standardizing to standard normal pp.77-79 "Check" 3.21.  We WILL learn to use table A. Moore doesn't separate out reading the z-table in the following; focus on just the z-table parts on first reading:  p. 79-83, Cum. proportion and  normal.  "Check" 3.22, 3. 23. "Backward" from prop. to z pp. 83-86.  Then we'll revisit and learn to deal with x's.  "Check" 3.24.
Read ahead:  Read Ch. 4 (Scatterplotts and correlation) to p. 104 Check p.112  4.14, 15, 16,   and NEXT  pp. 104-112 (correlation) Check 4.16 thru 4.22.  You do not have to be able to calculate r by hand.  You should be able to guess roughly at an r for a swarm of data; as p.108-9, and know and  be able to use facts 1-4, p. 107, and Cautions 1-4 pp. 108,110.
EXAM 1 still not finished.  Sorry!

Questions
on last HW? Day 10

Normal:  68-95-99.7% rule.  Standardization.  Table A, z to proportion.Day7
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New material:
Normal Distribution:  Using
Standard Normal Table "backward" (Day7 for details)  Normal curve templates --> you can use these and actually "count squares" to (approximately) check your work if you like.

Using Standardizing and the Standard Normal Table together to do more general problems.
Normal Practice
Handout
Review Standardizing
:  A "raw value" x is standardized by telling how many standard deviations above the mean it is.
Find z:  Subtract the mean from x.  Now you know how far "above" the mean x is, in "raw" units. (If it's below the mean, the number will be negative.)  Find how far this is in "standard deviations" by dividing by the standard deviation.
That's the z-score.

Standardizing:   A way of comparing an individual against its pack.
Comparing individuals from different packs, each relative to its own.
Removes "units of measurement" from the discussion.

Enables use of the standard normal table.

Examples: ("Classic IQ test", mean 110, s.d. 25)
85 is 1 s.d. below the mean.  Computation:  z = (85 110)/25 = (–25 raw points)/25 = –1 s.d. from mean.
145 is how many s.d.'s above the mean?
Computation: z = (145110)/ 25=  (35 raw points above mean)/25 = 1 2/5 = 1.4 s.d. above mean

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"What proportion"problems:  BPS5e pp. 79-83

•     Sketch a normal curve. Mark mean, 1, 2 s.d.'s.  Label with raw values, and z-values below (or above).
•     Mark end points for problem, roughly, and shade area desired.
•     Standardize end point(s).  Use standard normal table to find area. (Draw helper sketches if needed)
•     Check picture to see if it's plausible.
Example:  Proportion with scores between 100 and 145?  Table A (Excel)

x = 145 gives z = 1.4  (done before.)      Area to left of z = 1.4 is .9192  Proportion with scores below 145 is .9192
x = 100 gives z =  –.4                           Area to left of z = –.4 is  .3446  Proportion with scores below 100 is .3446
Desired area = Difference=  .5746;  about 57%.  Looks about right from picture.

or   P ( 100 < x < 145)  = P ( –.4 < z < 1.4) = P( z < 1.4) – P(z < –.4) = .9192 – .3446 = .5746
Read "Proportion of x's with 100 <x<145"  for P(100<x<145)

Proportion with scores above 145?  1 proportion with scores below.

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"Backward problems"  "What raw (x) value has area ___ to the left/right of it?"   BPS5e pp. 83-86.
Sketch  the curve, labeled with x values and z values, and the Area, roughly.
Restate (if needed) as "What z value has area A to the LEFT of it."
Look in body of table for the value closest to A.
Go to edge(s) of table to find what z that goes with.
Convert the z to an x:  z is the number of standard deviations above the mean.
Multiply z by the size of 1 standard deviation.  Now you have distance above the mean, measured in raw units.
Add the mean.  Now you have the "raw" value x. (You have "unstandardized" it.)
Example:  What x value has 10%  of the observations above it?  This is the same x as the one for:
What x value has 90% of the observations below (to the left of) it.

The table gives z = 1.28, approximately.  Table A (Excel)
The "Classic IQ test"score x= mean + z (s.d.) =  110 + 1.28 (25)=  110 + 32  = 142

Percentiles:  a "Classic IQ test" score of 142 has 90% of the scores at or below it.  142 is the 90th percentile.
Did it all.  Next time, exams back, questions on the above work, start Ch 4.

 Sievers home Math151-Sp12/Days11.htm 2pm 2/15/12
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