Checkpoint
- Course 2, Unit 7
Patterns in Chance
In each unit, there is a final lesson and Checkpoint that helps students
summarize the key ideas in the unit. The final Checkpoint will generally
be discussed in class, with the teacher facilitating the summarizing,
and students making notes in their Math Toolkits (teachers
may just refer to this as "notes") of any points they need to remember,
adding illustrative examples as needed. If your student is having difficulty
with any investigation in this unit, this Checkpoint and the accompanying
answers may help you recall the concepts involved, and give you the big
picture of what the entire unit is about. If your student has completed
the unit, then a version of this should be in his or her notes or toolkit.
Students should also have Technology Tips in their toolkits, which may
be useful for this unit.
Possible
Responses to Unit Summary Checkpoint
| In this unit, students explored the mathematics behind
waiting-time distributions. In the process, they discovered the Multiplication
Rule which can be used to find P(A and B)
when events A and B are independent. The bolded words
are vocabulary and concepts your student should be familiar with. |
| a. |
Write a general description
of a waiting-time distribution. Include how to construct the probability
distribution table, what the shape of the distribution looks like,
and ways to find the average waiting time.
- A waiting-time distribution (also known as a geometric
distribution) occurs in situations in which someone is
watching a sequence of independent trials and waiting for a
certain event to occur. For example, the trials could be a
person trying to shoot baskets and waiting for success. The
shooter may be successful on the first try, or the shooter
may have to wait for 10 shots for success to happen. It is
important that the trials are all independent; the basketball
shooter has the same chance of success on each attempt, no
matter how the last trial turned out.
- Suppose the probability of success is 40% on the first attempt,
then we would expect that 40% of the time you will be successful
on the first trial. Of those trials that do not produce a success
on the first try
(60% of
all trials),
you can expect that 40% of those (40%
of 60%) will produce a success on the second try. Of those trials
which do not produce success on the first two attempts (60% of
60%), you can expect that 40% of those (40% of 60% of 60%) will
produce a success on the third try. And so on. Of course, if
we are producing these trials by simulation, we will probably
not meet with success on the
first attempt exactly 40% of the time, but the same general
pattern will appear. The frequencies, and therefore, the probabilities,
drop as the waiting-time gets longer. The histogram illustrates
this.
- The formula for the probabilities in a waiting-time
distribution is P(x) = (1 - p)x - 1(p),
where x is the number of trials that had to be made to get
a success, or the length of the wait, and p is the probability
of success
on any one attempt. For the above example, where
the probability of a basket is 40%, the probability of having
to wait through 5 trials would be P(5) = (0.6)4(0.4).
This reflects the Multiplication Rule, since the probability
of having to wait through 5 trials would be the same as the
probability of (no and no, and no, and no, and yes), which
is P(no)4P(yes).
- To find the average wait time, or expected value, you
can find the mean of the frequency or probability distribution,
which
is
 or
1/p. This is also the balance point of the histogram. For example,
if P(yes) = 0.4, as in the basketball example and how
you try 100 different times to see how long it takes to make
a basket:
Wait
Time, x
(Number of Trials
to Produce a Success) |
Expected
Frequency, F(x) |
Total
Number of
Trials = xF(x) |
|
1
|
40
|
1(40)
or 40 trials
|
|
2
|
(0.6)(40)
= 24
|
2(24)
or 24 trials
|
|
3
|
(0.6)(0.6)(40)
= 14.4
|
3(14.4)
or 43.2
|
etc.
The longer the table, the better the approximation for the expected wait time,
since the number of trials can be any number from 1, 2, 3, ...
|
|
(expected
wait time) = (average number trials to produce
success) = (total number of trials)/(100)
= (sum of
this column)/100
If you stop after 3 lines of the table, you would estimate the
average
wait
time
is 1.07. If you stop the table after 10 entries, you would estimate that the
average
wait
time is 2.4 attempts. These are underestimates of the theoretical expected wait
time. (See below.)
|
Wait
Time, x
(Number of Trials
to Produce a Success) |
Probability |
Total "Trials" |
|
1
|
0.4
|
1(0.4)
|
|
2
|
(0.6)(0.4)
= 0.24
|
2(0.24)
|
|
3
|
(0.6)(0.6)(0.4)
= 0.144
|
3(0.144)
|
|
etc.
|
|
(average
wait time) = ∑ xP(x)
|
From the formula for the expected value of a waiting-time distribution, the
average number of trials is 1/p = 1/0.4 = 2.5.
|
| b. |
For what kinds of problems
and under what conditions should you use the Multiplication Rule
to calculate probabilities?
The Multiplication Rule is used when you want to find the probability
that two or more independent events all occur. |
If you would like to see specific problems from Course 2, Unit 7, a
link is provided to Examples of Tasks from
Course 2, Unit 7. If you are interested in following up on the Statistics
strand in general, then the Scope and
Sequence will help you see where different concepts are introduced.
On the Statistics page, you can read
an explanation of the main statistics concepts as they are developed
in all four courses.
|
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