1. For the probability table I gave in class find:

I will answer this Question later

  1. P(A)
  2. P(B)
  3. P(A intersect B)
  4. P(A U B) directly from the elementary events in the set and then using the addition rule:
    P(A) + P(B) - P(A intersect B)

2. The following is the probability distribution of the number of times a patient asks for a bed pan before receiving it on a unit.

X12345
P(X).6.2.1.07.03

Find the following:

  1. P(asking more than once) = 1 - .6 = .4
  2. P(asking more than twice given that the patient must ask more than once) =
    P({3,4,5} intersect {2,3,4,5}, given that {2,3,4,5} occured)=
    P({more than twice} intersect {more than once})/P(more than once) =
    P( {3,4,5} intersect {2,3,4,5})/ P({2,3,4,5}) =
    P({3,4,5})/P({2,3,4,5}) = .2/.4 = .5
  3. P(patients gets the pan within 3 requests) = P({1, 2, or 3})= .6 + .2 + .1 = .9
  4. P(asks less than twice or more than 4 times) P({1 or 5})= .6 + .03 = .63
I would conclude that care is poor because half of the patients who have to ask more than once, have to ask more than twice and 40% must ask more than once. (My concept (criterion) of quality is that few patients will have to ask more than once to get the bedpan). How could I tell care had improved?

3. Assume the probability of potassium levels are as follows


x[less than 3.5)[3.5 to 4.5)[4.5 to 5.5)[5.5 to 6.5)[6.5 and above
P(X).15.23.4.17.05

Note: [a , b ) is read " [include in the class interval the numbers from "a" up to, but not including, the number "b")

Find:

  1. P(of a level between [3.5 and 5.5)) = .23 + .4 = .63
  2. P(of a level equal to or greater than 4.5) = .4 + .17 + .05 = .62
  3. P(of a level between [4.5 and 6.5)) = .4 + .17 = .57
  4. Using the Addition Rule find P({of levels between 3.5 and 5.5} or {greater than or equal to 4.5}) . Show your work.
    A is the set {of levels between [3.5 to 5.5}
    B is the set {greater than or equal to 4.5}
    The intersection of A and B is the set { levels between [4.5 to 5.5}
    P(A) = .63 P(B) = .62 P(A intersect B) =.4 P(A U B) = P({greater than or equal to 3.5}) = P(A) + P(B) - P(A intersect B) = .63 + .62 - .4 = .85
  5. P( levels less than 4.5) =.15 + .23 + .38

4. For the distribution in home work problem 3 find:

  1. P(of levels less than 3.5 given that the values are less than 6.5) = .15/.95 = 3/19
  2. P(of levels between 3.5 and 5.5 given than the values are 3.5 or greater) = .63 / .85
  3. P(of levels 6.5 or greater given that the values are 4.5 or greater) = .05/.62

5. Three home care nurses have 25, 35, and 40 patients respectively. The severity is such that 5, 4, and 2 percent, respectively have to be rehospitalized within one month of discharge. One of their patients is readmitted to your unit. What is the probability that the patient came from the cases of nurse 1? nurse2? nurse3?

    Now we are ready to answer the questions posed in the problem:

  1. P(cared for by nurse A given readmitted) = P(cared for by nurse A and readmitted)/P(of readmission ) = .0125/.0345 = .3623
  2. P(cared for by nurse B given readmitted) = P(cared for by nurse B and readmitted)/P(of readmission ) = .014/.0345 = .4058
  3. P(cared for by nurse C given readmitted) = P(cared for by nurse C and readmitted)/P(of readmission ) = .008/.0345 = .2319
    This is the conditional distribution of nurse giving care given (or among) patients who were readmitted. Let's look at it expressed as a condititional probability distribution in a table.
    The Codnitional Distribution of Nurse given Readmits
    A.3623
    B.4058
    C.2319

6. A nurse wants to look at the effect of diagnosis on dietary intake of fluids. She observes patients with diseases A, B, and C and records the following intakes during waking hours. The probability distribution is as follows.
ABC
400 or less.20.04.10
400 to 800.05.15.15
more than 800.01.05.25

  1. What is the marginal probability for intake during the waking hours. Hint: how is the group distributed on intake?

    ABC
    400 or less.20.04.10.34
    400 to 800.05.15.15.35
    more than 800.01.05.25.31
    .26.24.501.0

  2. What is the marginal distribution of diseases. See above.
  3. What are the elementary events in the joint sample space.
    {(400 A) (400 B) (400 C) (800 A) (800 B) (800 C) (more A) (more B) (more C)}

  4. What is the conditional probability distribution for intake given disease A? given disease B? given disease C?
    Given AGiven BGiven C
    400 or less.20/.26 = .769.04/.24 = .1667.10/.50 = .2
    400 to 800.05/.26 = .19.15/.24 = .625.15/.50 = .3
    more than 800.01/.26 = .385.05/.24 = .208.25/.50 = .5
    1.01.01.0
  5. Looking at these three conditional distributions, is there evidence for independence of the variables: disease and intake.

    WEe use the definition of independence: P(A/B) = P(A) if A and B are independent.
    The conditionals for intake in disease A, B and C are not all equal to the marginal distribution of intake. Therefore the variables disease and intake are dependent.

  6. Find the probability that patients have (800 or less ) or (diseases B or C). Uses the addition rule. (Remember that you have to figure out what elementary events are in the intersection of these events and then knwo or be able to figure out probability of the intersection of them to use the addition rule.) After using the addition rule, write down the elementary events of interest and find the probability directly.
    Using Addition Rule: .69 + .74 - .44 = .99 Directly: .20 + .05 + .04 + .15 + .05 + .25 +.10 +.15 = .99
  7. Use the definition of independence, P(A intersect B) = P(A)P(B) which says "for any two events,
    the P(A intersect B) is equal to P(A) times P(B) if A is independent of B", to test whether having an intake of more than 800 cc is independent of disease category (A? B? C?).

7. On a unit marital problems among patients and their spouses occur in about 1/3 of the cases. The need for dependent care from the spouse after discharge occurs about 60% of the time. Assuming that marital problems and the need for dependent care are independent find the joint distribution of these two variables in the patient population. (Write the joint probability distribution.)

need dependent careno dependent care
marital problems.20.133.33
no marital problems.40.267.66
.60.401.0