| Hand in
(All D&V) standardizing: Ch6 p. 99: Sketch each Normal model and label its axis with both the "real/raw" values and the "z" values. Mark the observations on the pictures, do questions. 5 temperatures 6 placement exams = = = = = = = = = = = = = Table use: Always sketch the model first, mark the area you are looking for! Find the answers using Table Z, Appendix p. A-30. Check your answers with one of the Technology Normal tools (see above) p.101 #20, 22 z's (Note: 22d finds what numbers from the 5-number summary?) Raw data problems. Do unstarred parts with table Z. "Backward" parts are marked with *--do them with a technology tool (see above) p.102, 25 Cholesterol a, b, c, d*, e* 26 Tires a, b, c, d*, e* 28 Body Temperatures: a, b, c* Also: I have a theory as to where the "wrong" number 98.6F came from. Early work on temperatures all took place in Europe. Convert 98.6F and 98.2F to Celsius (subtract 32, and divide by 1.8). What's my theory? |
Read,
to discuss |
Optional
(more
practice) More table practice: z's: p.101 #19, 21 p.102, 27 Kindergarten a, b*, c* p.109, 25 BeQuick a,b,c,d,e*,f * |
Examples: (Wechsler 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 = (145
– 110)/
25=
(35 raw points above mean)/25
=
1
2/5 = 1.4 s.d. above mean

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"What proportion"problems:
(D&V p.90-2: "....I", "More..." question 1) http://www.whfreeman.com/scc/

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)
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"Backward problems" "What
raw (x) value has area ___ to the
left/right
of it?" (D&V p.90-2: "....II")
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")
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.
The Wechsler score x=
mean
+ z (s.d.) = 110
+ 1.28 (25)= 110
+ 32 = 142
Percentiles: a Wechsler score of
142 has 90% of the scores at or below it. 142 is the 90th
percentile.
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