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:
| Hand in
>>standardizing: I recommend drawing the two "rulers"--raw and z-score. p. 56 1.56 SAT/ACT p. 65 1.64 (cf. batting avgs) >>table use: Always sketch the distribution first, mark the area you are looking for! (Use http://www.whfreeman.com/scc/ to check.) p. 64 1.68 a and b.Pregnancies Also (with 1.68), What proportion of pregnancies last 310 days or more? (see below **) p. 61 1.58 (locomotive adhesion, 2 dist's) p. 66 1.69 (Stanford-Binet, "superior") >>"Backward Normal"Always sketch a normal curve first, roughly mark the proportion=area you are given. p. 76 1.89 (soldiers' heads) p.64 1.68 c (pregnancy) p. 66 1.70 (z-scores of quartiles) |
Read, to discuss | Optional (more practice)
"Backward"
"Backward"
|
Dear Abby: You wrote in your column that a woman is pregnant for 266 days. Who said so? I carried my baby for ten months and five days, and there is no doubt about it because I know the exact date my baby was conceived. My husband is in the Navy and it couldn't have possibly been conceived any other time because I saw him only once for an hour, and I didn't see him again until the day before the baby was born. I don't drink or run around, and there is no way this baby isn't his, so please print a retraction about that 266-day carrying time because otherwise I am in a lot of trouble.Abby's answer was consoling and gracious but not very statistical:
San Diego Reader
Dear Reader: The average gestation period is 266 days. Some babies come early. Others come late. Yours was late.
The question here is not whether the baby was late. That
fact is already known. At issue is the credibility of the length of the
delay. Ten months and five days is approximately 310 days, which means
that the pregnancy exceeded the norm by 44 days. [How unusual is that?]
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: 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:
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?"
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) = 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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