a question on bio-statistics
nurse statistics
following question has an answer.
The probability of getting pregnant while using condom varies with
age, ethnicity, etc.
But assume it is 20%. That is, if 100 women use condom in all sex
relations during one year, then in the end of the year 20 of them are
going to be pregnant.
Now my question is: if the 100 women keep doing the same for five
years, how many of the 100 are going to have a pregnancy during these
five years?
Exists a function p(years)=?
If the probability in one year is 20%, then in n years is?
And is there a function depending on years and on probability?
p(probability,years)=?
I thank all your help.
Chip Eastham
> I am a nurse and do not know much about mathematics. I hope the
> following question has an answer.
>
> The probability of getting pregnant while using condom varies with
> age, ethnicity, etc.
> But assume it is 20%. That is, if 100 women use condom in all sex
> relations during one year, then in the end of the year 20 of them are
> going to be pregnant.
>
The simplest approach assumes the chances
in different years are independent, e.g.
that the probability of pregnancy in two
different years is the product of the
probabilities of pregnancy in each year,
and that the individuals' probabilities
are also independent. With those
assumptions we have a binary distribution
of individuals having at least one
pregnancy after n years:
Pr( k pregnancies in n years )
= C(n,k) (0.2)^k (0.8)^(n-k)
Here C(n,k) is the binomial coefficient,
read "n choose k", equal to n!/(k!(n-k)!).
The hat ^ represents "raised to the power".
> Now my question is: if the 100 women keep
> doing the same for five years, how many of
> the 100 are going to have a pregnancy
> during these five years?
>
> Exists a function p(years)=?
> If the probability in one year is 20%, then in n years is?
>
> And is there a function depending on years and on probability?
> p(probability,years)=?
>
> I thank all your help.
Let's apply the formula to estimate the
number of individuals out of a sample of
100 who become pregnant at least once in
five years. The simplest way is to count
the number expected to have zero "gravida"
in those five years and subtract from 100:
Expected(zero gravida) =
100 * C(5,0) * (0.2)^0 * (0.8)^5
= 100 * (0.8)^5
~ 32.8
Rounding off to avoid fractional nurses,
we would expect 33 of the 100 to avoid
pregnancy during the five years. Thus
about 67 would be expected at least one
pregnancy in those five years.
Of course the simple statistical model
is not entirely realistic: pregnancy
events in one year are likely to affect
the chances of pregnancy in another year.
But this is how the math works for such
an approximation.
regards, chip
Ray Vickson
> I am a nurse and do not know much about mathematics. I hope the
> following question has an answer.
>
> The probability of getting pregnant while using condom varies with
> age, ethnicity, etc.
> But assume it is 20%. That is, if 100 women use condom in all sex
> relations during one year, then in the end of the year 20 of them are
> going to be pregnant.
Not quite. The number of women (out of 100) who will get pregnant in
one year is, itself, a random quantity that will vary from year-to-
year or from one group of 100 to another such group. However, when
averaged over all such groups or over all years, the _average_ number
getting pregnant is 20. The actual number in any one group or in any
one year is a random quantity that follows the so-called _Binomial
distribution_. Its mean is 20, its "variance" is 16 and its standard
deviation (the square root of variance) is 4. See
http://en.wikipedia.org/wiki/Binomial_distribution for more on the
binomial distribution. In n years the number getting pregnant
(assuming that multiple pregnancies is allowed for the same person) is
the binomial distribution with mean 20*n and standard deviation 4*sqrt
(n) (sqrt = square root). In fact, this is the same as for a group of
100*n women for one year; the same formulas apply.
To see what is happening in a simple case, consider just two women, W1
and W2. In one year each can be N (not pregnant) or P (pregnant). Here
is a little table of the possibilities, together with their
probabilities:
women/state probability
W1(P) W2(P) (0.2)(0.2) = 0.04
W1(P) W2(N) (0.2)(0.8) = 0.16
W1(N) W2(P) (0.8)(0.2) = 0.16
W1(N) W2(N) (0.8)(0.8) = 0.64
---------
Total 1.00
The probabilities for numbers of pregnancies are: P{0} = 0.64, P{1} =
0.16 + 0.16 = 0.32 and P{2} = 0.04. The average number pregnant is 0*P
{0} + 1*P{1} + 2*P{2} = 0.32 + 2(0.04) = 0.4. Note that this is = 2*
(0.2) = 2*P{woman get pregnant}. For a group of 100 women the
calculations would be similar but much more tedious. In general, the
probability that exactly k women are pregnant is (0.2)^k * (0.8)^(100
- k) * Num(100,k), where Num(100,k) = number of distinct ways of
having k "P's" and 100 - k "N's". There is a formula for Num(100,k),
but it is not very useful in this case (essentially because the number
100 is large), but you can see, in principle, how the counting could
be done. If you have access to EXCEL (or comparable open-source
spreadsheets) you can get the probabilities easily just by going into
the appropriate menu and pulling up the "binomial distribution" icon.
R.G. Vickson
Robert Israel
> On Jul 4, 7:54=A0am, nurse statistics <nurse.st...@gmail.com> wrote:
> > I am a nurse and do not know much about mathematics. I hope the
> > following question has an answer.
> >
> > The probability of getting pregnant while using condom varies with
> > age, ethnicity, etc.
> > But assume it is 20%. That is, if 100 women use condom in all sex
> > relations during one year, then in the end of the year 20 of them are
> > going to be pregnant.
>
> Not quite. The number of women (out of 100) who will get pregnant in
> one year is, itself, a random quantity that will vary from year-to-
> year or from one group of 100 to another such group. However, when
> averaged over all such groups or over all years, the _average_ number
> getting pregnant is 20. The actual number in any one group or in any
> one year is a random quantity that follows the so-called _Binomial
> distribution_. Its mean is 20, its "variance" is 16 and its standard
> deviation (the square root of variance) is 4. See
> http://en.wikipedia.org/wiki/Binomial_distribution for more on the
> binomial distribution. In n years the number getting pregnant
> (assuming that multiple pregnancies is allowed for the same person) is
> the binomial distribution with mean 20*n and standard deviation 4*sqrt
> 100*n women for one year; the same formulas apply.
Except that it's more complicated than that. Once a woman is pregnant, if
nature takes its normal course she doesn't get pregnant again for about
9 months, followed by a time of reduced fertility that can vary widely.
Fortunately, for what seems to be the actual question (rather than the one
you're answering) that's not really relevant.
> To see what is happening in a simple case, consider just two women, W1
> and W2. In one year each can be N (not pregnant) or P (pregnant). Here
> is a little table of the possibilities, together with their
> probabilities:
>
> women/state probability
> W1(P) W2(P) (0.2)(0.2) =3D 0.04
> W1(P) W2(N) (0.2)(0.8) =3D 0.16
> W1(N) W2(P) (0.8)(0.2) =3D 0.16
> W1(N) W2(N) (0.8)(0.8) =3D 0.64
> ---------
> Total 1.00
>
> The probabilities for numbers of pregnancies are: P{0} =3D 0.64, P{1} =3D
> 0.16 + 0.16 =3D 0.32 and P{2} =3D 0.04. The average number pregnant is 0*P
> {0} + 1*P{1} + 2*P{2} =3D 0.32 + 2(0.04) =3D 0.4. Note that this is =3D 2*
> (0.2) =3D 2*P{woman get pregnant}. For a group of 100 women the
> calculations would be similar but much more tedious. In general, the
> probability that exactly k women are pregnant is (0.2)^k * (0.8)^(100
> - k) * Num(100,k), where Num(100,k) =3D number of distinct ways of
> having k "P's" and 100 - k "N's". There is a formula for Num(100,k),
> but it is not very useful in this case (essentially because the number
> 100 is large), but you can see, in principle, how the counting could
> be done. If you have access to EXCEL (or comparable open-source
> spreadsheets) you can get the probabilities easily just by going into
> the appropriate menu and pulling up the "binomial distribution" icon.
>
> R.G. Vickson
>
>
>
> >
> > Now my question is: if the 100 women keep doing the same for five
> > years, how many of the 100 are going to have a pregnancy during these
> > five years?
For a given woman, the probability of not getting pregnant in any one year
is 1 - 0.2 = 0.8, so the probability of not getting pregnant in 5 years is
.08^5 = .32768, so the probability of getting pregnant at least once in
5 years is 1 - .32768 = .67232. Again, we don't know how many of the 100
women will get pregnant at least once in 5 years, but the expected number
is 100 * .67232 = 67.232.
> > Exists a function p(years)=3D?
> > If the probability in one year is 20%, then in n years is?
> >
> > And is there a function depending on years and on probability?
> > p(probability,years)=3D?
> >
> > I thank all your help.
>
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada