Math 151 , Spring 2005, Day 26 Wed. April 6 Hit reload  After class

Exams still not finished.  Friday for sure!
Day 26 (Mon. April 6): Reading: Continue D&V Part IV: Ch.15 thru p.  291 (then Ch. 18 &on. Start 18 now!) ActivStats is very good for part IV--Ch11"Randomness" shows Law of Large Numbers as D&V express it. Ch14, 15"Intuitive Probability"&"Probability Rules" correspond well with the text and present very good examples.
Hand in
Continuous sample spaces: 
Do the questions A and B in webpage below, with the Tables for simple models (densities) handout
Normal model: Restate each problem using:  "The probability that x is..." ~ ~ "The proportion of the population of x's that ...." and use your old techniques.
p. 352 #21 a, b only. Pregnancy Find the solutions, then Restate a and b from proportion questions to probability questions:  "What is the probability that a pregnancy chosen at random will last... ... " and "If the chances of a random pregnancy lasting longer than k days is.... ..., then what is k?"
p. 265 #28a only Rivets
p. 353 #33a only IQ's,  #34a only Milk

Probability, Ch. 15. p. 299
3 Homes, 5 Amenities  (Use Step-by-step p. 290 to pattern on)

A.  Newly assigned after class, for Day 26: On separate page: Use the "Probability" Applet   at http://bcs.whfreeman.com/scc  to simulate tossing a penny 25 times.  Write down h, the number of heads, and p-hat, the proportion of heads you got (p-hat = h/25).  Reset, and repeat, till you have a total of 10 simulations.  Make a dotplot (by hand; see D&W p. 36) of your  10 p-hats, using an axis marked with .28, .32, .36, .40, .44, .48, .52, .56, .60, .64 etc. (add numbers at the ends if you need them).  Bring your list and dotplot to class to hand in and to share.  (Be glad I didn't have you flip a real penny.)

On separate paper: Start the rest; read them all.  Do what you can.  Will be assigned next time.
8 Pets  Make a 2-way table using the counts given.  Use it to find the conditional probabilities.
9 Health, 10 Death penalty  Mixture of conditional/not conditional problems.  Read carefully!
23 Health, independence?
22 Pets, independence
29 a Absenteeism

Read,
  to 
discuss 
Optional 
Homework questions?  Day 25
&&Continuous sample space:  If the sample space is an interval of values (or the whole line), the possible outcomes are "x" or "y" values in the interval.  The way we assign probabilities to events is with a density (Day 7). (Remember density curves were idealizations of histograms--of repeating the "experiment" many many times.)
Area represents proportion-->> Area represents probability.

  P(a < x < b) = the probability that the outcome x is between a and b
                      is the area under the model's density curve, between a and b.
                      is the proportion of x's which would come up between a and b if we did the phenomenon a zillion times.
We declare P(a) = 0 (In a continuous model, getting precisely a is utterly unlikely; can't even measure that well),
       so P(a < x < b) = P(a < x < b)

Review ""Tables for simple models (densities)"" HW day 7, restating these parts as probability questions:
    (Copies of the HW handout are outside my door if you can't find yours.)
Change language from "description of a population of data"  or "area between/above/below" to
   "pick an individual at random the population, call the value x"
A. ("Uniform")  Spin the spinner once.  x = number the spinner points to.
a) (example)  The probability that the spinner points to a number less than .6 = P( x < .6) = area to left of .6 = .6.
b) P (.2 < x < .6) = ?   Say it in words: ?
c) For what c is there probability .4 of being greater than c ?      (In notation: P(x > c) = .4.  Find c)
B.  y = (number you get from) the sum of two spinners. ("Triangular")
a) The probability that the sum is a number less than .6  =  P(       ?        ) =.18
b) P(y > 1.6) =  ?     P(y < 1.6)  =         P (y < 1) =   ?             P( 1 < y < 1.6) =  ?   Say each in probability words.
c)  P(c > x) = .08.  Find c:  ?   Say in words.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Our most important probability model: NORMAL Model family.  Same techniques as before, only we ask "probability that one chosen at random..." instead of "proportion of all..."  Review Normal techniques: Day 8, Day 9
 Take a random sample of size 1 from a population which is N(110, 25) = =
Give an individual adult, chosen at random, the Wechsler test, which has a normal distribution, mean 110, s.d. 25.   x is her score on the test.
Find P(100 < x < 140), prob. that the  individual chosen at random gets between 100 and 140.  Same as: of all  individuals, fraction who score between 100 and 140.  Work is on Day 9, what proportion.
= = = = = = = = = = = = = = = = = =
Back to Chapter 16, Probability rules
Equally likely probabilities:  If there are k possible outcomes, and each is equally likely, each has probability 1/k (Dice, coin, etc. Plenty of things are not equally likely.)
(Remember:)  If A and B are disjoint events, then P( A or B) = P(A) + P(B)
  If A and B can happen together,  P(A and B) is not 0.
    Then P(A or B) = P(A) + P(B) - P(A and B).  "General addition rule."
     Use Venn diagram to see.   Example
A thousand people are interviewed by the census bureau, and the results tabulated in this two way table.
Working Status vs. Sex.
Women Men Total
In Labor Force 350 450 800
Not in Labor Force 150 50 200
Total 500 500 1000
 
Restated as proportions of the whole: 
Women Men Total
In Labor Force .35 .45 .80
Not in Labor Force .15 .05 .20
Total .50 .50 1.00
 
Now choose one person at random from this group.  A = Person is In Labor Force , B = Person is a woman
  P(A and B) = Person is in the labor force and a woman = .35
 P(A or B) = Person is in the labor forceor a woman = .80 + .50 - .35 = .55

Two  variables:  Restate conditional frequency distributions as conditional probabilities.  Recall Day 3
   P(B| A):  Probability of B "given" A:  Probability that B happensconditional on knowing that A happens.

When you write or see percents, be clear what is on the  bottom of the fraction (even if it takes longer to say)!!.
Start here Friday
Marginal distribution:  Distribution of one variable, ignoring/summingover the other.
Working Status
In Labor Force 800 80%
Not in Labor Force 200 20%
Total 1000 100%
Sex
Women Men Total
500 500 1000
50% 50% 100%

   P ( Woman)=  50% = .5,   P( In Labor Force) = 80% = .8.
Conditional distribution:  Distribution of one variable, with the individuals being only those which satisfy a condition in the other variable.
 Choose a person:  Conditional distribution of working status given that  a woman was chosen. Dist. of working status given that a man was chosen.
Women Men Total
In Labor Force 350/500 = 70% 450/500 = 90% 80%
Not in Labor Force 150/500 = 30% 50/500 = 10% 20%
Total 500/500=100% 500/500=100% 100%

   P( In Labor Force | Woman) = 350/500 = .7    P (In Labor Force | Man) = 450/500 = .9
Given that the person chosen is in the labor force, conditional distribution as to sex.
    Given that the person chosen  is not in the labor force, conditional distribution as to sex.
           "Row %s"--rows add to 100%:  "conditional distributions of sex by working status."
Women Men Total
In Labor Force 350/800 = 43.8% 450/800 = 56.2% 800/800=100%
Not in Labor Force 150/200 = 75% 50/200 = 25% 200/200=100%
Total 50% 50% 100%
P( Woman | In Labor Force ) = 350/800 = .438     P (Man | In Labor Force) = 450/800 = .562 (not the same as above)

Another formula:   P(B| A) = P(Band A)/P( A)
  P( Woman | In Labor Force ) = P( Woman and In Labor Force )/ P( In Labor Force )
                                                                          = (350/1000) ÷ (800/1000) = .35/.80 = 350/800 = .438
Independence (definition)  A and B are independent if - and only if-   P(B| A)  = P(B)
   Knowing A is true (or not) makes no difference to the probability of B.
   A being true (or not) does not change the chances for B.

Now if A and B are independent, then  P(B| A)  = P(B) .  But P(B| A) = P(Band A)/P( A) .
  So P(B) = P(Band A)/P( A) .  Multiply both sides by P( A). Get
    P(Band A) = P(B) × P( AThis was our Rule 5 from p. 275 (Day 23, bottom)

IF Sex and Working Status were independent, the table would look like this: (or close to it)
 
Working Status vs. Sex.
Women Men Total
In Labor Force 350 400 450 400 800
Not in Labor Force 150 100 50 100 200
Total 500 500 1000
 
Restated as proportions of the whole: 
Women Men Total
In Labor Force .40 .40 .80
Not in Labor Force .10 .10 .20
Total .50 .50 1.00
 
Now to Chapter 18, probability applied to simple random samples!


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there will be none or only one from  0, 1, 5, 9.  At least one and probably more 7's. A"bias" against 0,1,5,9, and toward 7.