Database Reference
In-Depth Information
speciied value, X. We also depicted the reverse or inverse problem of specifying
a probability/percentage of time a result comes out less than “X,” and determining
what the value of X is.
We need to add one simple aspect to what we have already discussed. Up to now,
we've been inquiring about probabilities for an individual user. (When we make a
statement such as, “95% of the users will have a time between 63 and 75 seconds, we
are referring to 95% of the
individual
users.”) But, what if we wish to inquire about
the
mean
of a speciic number of users from the same population? That is, how do
things change if we wish to ask about the behavior of the
mean of
n
users' completion
times
? For example, suppose that we ask, “If we have a sample of four users inde-
pendently (i.e., separately) completing a task, what is the probability that the
mean
completion time exceeds some value?”
2
To answer the question, let's go back to the problem posed in the previous section
in which the average time of hooking everything up for the Internet installation (for
the whole population of users and potential users) was 160 seconds and the standard
deviation was 20 seconds. We showed that the percentage of individual users who
would require a completion time less than 190 seconds was 93.32%.
To answer the question about the
mean time of
4
users
being less than 190, we
use the same NORMDIST(X, μ, σ, 1) command, but adjust the standard deviation to
relect that we are now asking about the behavior of a mean of a sample of “
n
” (4 in
the speciic question), and not the behavior (i.e., time) for just an individual person.
It certainly is a different question to ask what the time will be
on average
for four
people and what the time will be for an individual (one) person.
The adjustment is simply to replace the value of the standard deviation, σ,
which is known to be 20, by what we can call the “adjusted standard deviation”
(oficially called “the standard deviation of the mean”) which is the original stan-
dard deviation (i.e., for one person/time), divided by the
positive square root of
the sample size comprising the mean
—here, the sample size equals 4. So, the
adjusted standard deviation is 20 divided by the positive square root of 4, which
equals, of course, 2.
The resulting adjusted standard deviation is, hence, 20/2 = 10. Now we use the
same Excel command we used earlier, but with a “10” instead of a “20.” We repeat
the earlier command for reference, and then the new, revised command:
=NORMDIST(190
,
160
,
20
,
1) =0
.
9332or93
.
32%
=NORMDIST(190
,
160
,
10
,1) =0
.
9987or99
.
87%
In other words, one user's time has a 93.32% chance of being less than 190
seconds, but the average of four users' times has a (gigantic!) 99.87% chance of
being less than 190 seconds.
2
Any time that we take a sample, the reader should assume that it is a “random sample,” unless
indicated otherwise. A random sample is one in which every possible sample of the “population” from
which the sample is drawn has an equal chance of occurring.
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