Day 5 Math251

MATH 251, Probability and Statistics I, Fall 2007, Day 5After class. Hit reload!

Meet in Mac 101 lab Wednesday for big SPSS intro. Bring a disk or usb, book.
Day 5, Monday Sept. 3 [stuff in brackets all use the table "backwards." Try the ones in red, but put them on separate paper, to be part of Day 7.]
Reading: IPS 1.3 Density curves 64-69 & Normal distribution, pp.64-80.
Read ahead: Normal quantile plots, 80--84; then Ch. 2

Hand in:
N.b. "cf." is short for "compare to" in reference speak. From Latin "confer." N.b.="nota bene", note well.

Normal density p. 84ff. Always sketch the curve, mark the area(s) you need.
1.84 (sketch) 1.86 (pregnancies, rule)
1.87 (horse preg's , rule)
1.88 (variability in poll, rule)
1.90 Use Applet to find Quartiles in Normal
1.89 (heights, compare)
1.99 (SAT/ACT, compare)

1.92, [1.94] (standard normal table)
- - - - - - - - - - - -
The rest of these on separate paper as part of Day 7: Do all of these using the Normal Density Applet http://www.whfreeman.com/ips5e/ to get answers. Leave space to do the calculations to get the answers from Table A. Always sketch the curve, mark the area(s) you need.
Non-standard normal, table problems:
    X is normal with mean 3 and s.d. 2. Find:
    The proportion with X<1.5. 1.5 < X < 4.5.
    [The x for which 30% of the observations are
              smaller than it. The 30th percentile. ]
1.96, 1.97 (Wechsler WAIS)
1.112 (osteoporosis)
1.113 (tails)
1.115 a, b, [c] (pregnancy)
p. 98, 1.141 d. 78's are Raw scores. I don't know why the first sentence is here--seems irrelevant to the question.
B. What percent of pregnancies last 310 days?
    (cf. 1.115, see quote below)

p. 90, 1.116 (compute quartiles, pregnancy. cf. 1.90)
1.101 (ACT equiv. SAT)
1.106 (SAT goal)

Read, discuss
1.78
Normal density

1.93, [ 95] (more table practice)

Optional

Normal density
1.85 Use Normal Density Applet to check on Rule
1.98 (z-scores, more practice)

Use Normal Density Applet http://www.whfreeman.com/ips5e/ to check on all your Normal calculations!

X is normal with mean 3 and s.d. 2. [Find the x for which 80% of the observations are smaller than it. The 80th percentile.]

[(Following 1.116, do 1,117, 1.118, IQR and outliers in Normal]

[In 1973] the following item appeared in Dear Abby's column:

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.
San Diego Reader

Abby's answer was consoling and gracious but not very statistical:

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?]

Meet in Mac 101 lab Wednesday for big SPSS intro. Bring a disk or usb, book.
Math251@wells.edu: Use to communicate questions, answers, look for HW buddies...
Handout: solutions to B, and 1.141 ab, Day 4 . Work on them. Will be on in-class quiz after a while.
Density handout solutions
Other HW questions?
Densities--abstraction from histogram. "Model" Day 4
Median, mean, percentiles, standard deviation are defined for a density model in analogy to those for a histogram.
-- median has half of area below and half above.
-- mean is balance point. On the long-tail side of median if distribution is skewed. Same as median if symmetric.
--First quartile has 1/4 of area below, 3/4 above. Etc. for others.
--Greek labels "mu" for mean and "sigma" for std. dev. of a Density.
"Normal" Density :("Gaussian", "Bell-shaped") Normal Density Applet http://www.whfreeman.com/ips5e/

What "mechanism"? &&Thing measured is the result of many small independent influences.
Family: all the same basic shape, except for measurement units.
Mean gives middle, standard deviation gives spread.
Knowing parameters mean (µ "mu") and standard deviation ("sigma") tells you everything. "N(mean, s.d.)"
Symmetric. Mean at middle. Curvature changes at 1 standard deviation on either side of mean.
"Standard normal": mean = 0, s.d. = 1 Standardized : how many "s.d.'s from the mean".
"Middle s. d." && A center region one standard deviation wide spans about 38% (Not in IPS)
68-95-99.7 rule: 68% within 1 s.d. of mean, 95% within 2 s.d. of mean, 99.7% within 3 (roughly).

higher

Back-of-the-envelope: Sketch the curve, mark mean, + 1, +2, +3 s.d.'s-- in real units.

Example:

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: Psychology test "W" scores are approximately N(110, 25)
   A score of   85 is 1 s.d. below the mean. Computation: z = (85 – 110)/25 = (–25 raw points)/25 = –1 s.d. from mean.
           (About the 16th percentile--16% get scores < 85)
   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

Start here Friday (SPSS Wed.)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ First standard normal table use, then with "real" values~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Standard Normal N(0, 1). Our tables give area to the left of a z value. Table A, front flyleaf
Using standard normal table: p. 77
       z | .00     .01     .02 .....
      ...|
     1.4 | .9192   .9207   .9222 ....
   P(Z < 1.40) = .9192,   P(Z < 1.41) = .9207 P(Z < 1.42) = .9222.
                                              ?z has more than 2 dec. places? Round to 2.

    Sketch the density, mark the area you're looking for.
    Figure out how to get it using areas to the left of one or more z-values.
        Think cutting up paper bell-curves. (Remember whole area is 1.) Like handout.

Example: Proportion of observations between 0.5 and 1.4 P(0.5 < Z <1.4) =
Proportion of observations below 1.4 minus Proportion of observations below 0.5
P (Z < 1.4) - P(Z < 0.5) = .9192 - .6915 = .2277

. bell curves. Use 202x515 pixels to print.
Example: Proportion of observations above 0.5,    P( Z > 0.5) =
                ONE minus proportion of observations below 0.5,   1 - P(Z < 0.5) = 1-.6915 = .3085
. Reading table "backward":
What z value has area ..... to the left/right of it?
        Sketch 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.
Example: "What z value has 10% of the observations above it?" This is the same z as the one for:
        "What z value has 90% of the observations below (to the left of) it." (What z is the 90th percentile.)

        Find in the table .8997 and .9015 -- .9000, our number, is between them.
                    .8997 is a little closer to.9000, so use it.
           For .8997, the z value is 1.28.   1.28 is the 90th percentile.
            1.28 has 10% of the observations above it.

Real/Raw data:
"What proportion"problems:

Sketch a normal curve. Mark mean, 1, 2 s.d.'s. Label with raw values, and z-values below.
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?

x = 145 gives z = 1.4 (done above.)      Area to left of z = 1.4 is .9192
x = 100 gives z = –.4                           Area to left of z = –.4 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)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
"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: "W" test: 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 "W" score x= mean + z (s.d.) = 110 + 1.28 (25)= 110 + 32 = 142
Percentiles: a "W" score of 142 has 90% of the scores at or below it. 142 is the 90th percentile.

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