Biomedical Engineering Reference
In-Depth Information
tion of 28.5 counts. (a) What is the probability of observing 800 or fewer counts in a
given minute? (b) What is the probability of observing 850 or more counts in 1 min?
(c) What is the probability of observing 800 to 850 counts in a minute? (d) What is
the symmetric range of values about the mean number of counts within which 90%
of the 1-min observations are expected to fall?
Solution
The normal distribution is indistinguishable from the binomial and Poisson distri-
butions within the precision given in the problem. Whereas the desired quantities
would be tedious to calculate by using either of the latter two, they are readily ob-
tained for the normal distribution.
(a) We use as estimates of the true mean and standard deviation
813 and
σ
= 28.5. Let
x
represent the number of counts observed in 1 min. We have for the
standard normal random variable, from Eq. (11.39),
z
µ
=
(
x
- 813)/28.5. The probabil-
ity that
x
has a value less than or equal to 800 is the same as the probability that
z
is
less than or equal to (800 - 813)/28.5
=
=
-0.456. Interpolating in Table 11.1, we find
that
P
(
x
≤ 800) =
P
(
z
≤ -0.456) = 0.324.
(b) For
x
=
850,
z
=
(850 - 813)/28.5
=
1.30. Whereas Table 11.1 gives values
P
(
z
≤
z
0
), we are asked here for a complementary value,
P
(
z
≥
z
0
)
=
1-
P
(
z
≤
z
0
). Thus,
P
(
x
0.097.
(c) It follows from (a) and (b) that the probability that
x
lies between 800 and 850
is equal to the area under the standard normal distribution in the interval 0.456 <
z
<
1.30. Using Table 11.1, we find that
≥
850)
=
P
(
z
≥
1.30)
=
1-
P
(
z
≤
1.30)
=
1 - 0.9032
=
P
(800 <
x
< 850) =
P
(-0.456 <
z
< 1.30)
(11.42)
=
P
(
z
< 1.30) -
P
(
z
< -0.456)
(11.43)
= 0.903 - 0.324 = 0.579.
(11.44)
As a check, the sum of the answers to (a), (b), and (c) must add to give unity (except
for possible roundoff ): 0.324 + 0.097 + 0.579
1.000.
(d) With respect to the standard normal curve, 90% of the area in the symmetric
interval
=
±
z
about
z
=
0 corresponds to the value of
z
for which
P
(
z
≤
z
0
)
=
0.9500.
From Table 11.1 we see that this occurs when
z
0
=
1.645. Equation (11.39) implies
that the corresponding interval in
x
is
±
1.645 standard deviations from the mean
µ
.
Thus, 90% of the values observed for
x
are expected to fall within the range 813
±
1.645(28.5) = 813 ± 46.9.
Table 11.2 lists some useful values of one-tail areas complementary to the areas
given in Table 11.1. To illustrate its use, the answer to part (b) of the last example
can be gotten by interpolating in Table 11.2 with
k
α
= 1.30
, giving
P
(
x
≥ 850) =
0.0975
. Also, the second entry in the table shows that one-half of the area under a
normal distribution lies within the interval
±
0.675
σ
about the mean.
