AMS 311 (Fall, 2010)
Joe Mitchell
PROBABILITY THEORY
Homework Set # 9 – Solution Notes
(1).
(20 points)
The AMS department receives, on average, three requests per day for students to sign into the major.
We do not know the probability distribution for the number,
X
i
, of students who sign into the AMS major on day
i
.
Let
X
i
be the number of students who sign into the AMS major on day
i
. We know that
E
(
X
i
) = 3.
(a). Let
p
be the probability that Fve or more students sign into the AMS major on Monday. Give the best
guaranteed
estimate you can for the probability
p
. (What inequality are you using?)
(We think of Monday as “day
i
”.) We use Markov’s inequality to get a bound on
p
:
0
≤
p
=
P
(
X
i
≥
5)
≤
E
(
X
i
)
5
=
3
5
= 0
.
6
(b). ±or the next three parts ((b), (c), (d)) assume that we also know that the variance,
var
(
X
i
)
, is 9. Try now to
use an appropriate Chebyshev inequality to give an improved pair of bounds (upper and lower) on the probability
p
.
Now we know that
var
(
X
i
) = 9. So we apply the onesided Chebyshev inequality, with
μ
=
E
(
X
i
) = 3 and
a
= 2:
0
≤
P
(
X
i
≥
5) =
P
(
X
i
≥
3 + 2)
≤
9
9 + 2
2
=
9
13
= 0
.
692
Note that, in fact, in this case the Chebyshev inequality gives a
weaker
bound than the Markov inequality did in (a).
(This answers a question that was asked in lecture!)
(You could have used a 2sided Chebyshev, which would result in a weaker bound as follows:
P
(
X
i
≥
5)
≤
P
(
X
i
≥
5 or
X
i
≤
1) =
P
(

X
i

3
 ≥
2)
≤
9
2
2
=
9
4
= 2
.
25. In fact, this is such a weak bound that it is useless, since 2
.
25
>
1.)
(c). Give a Central Limit Theorem estimate for the probability
q
that more than 75 students sign into the AMS
major during this month (December, which has 31 days, and we consider each day to be like any other day).
Let
X
i
be the number of students signing into the AMS major on December
i
, for
i
= 1
,...,
31. Let
X
=
∑
31
i
=1
X
i
denote the total number of students signing into AMS major in December. Then,
E
(
X
) = 31
·
E
(
X
i
) = 31
·
3 = 93 and
var
(
X
) = 31
·
var
(
X
i
) = 279 (using the fact that the
X
i
’s are independent). Then, using the Central Limit Theorem
estimate, we get
q
=
P
(
X >
75) =
P
(
X

93
√
279
>
75

93
√
279
)
≈
P
(
Z >

1
.
0776)
≈
Φ(1
.
08) =
.
8599
(d). Use an inequality to get the best bounds you can on the probability
q
estimated in part (c).
We can apply the onesided Chebyshev inequality to get We want to estimate
P
(
X >
75) = 1

P
(
X
≤
75). Now,
P
(
X
≤
75) =
P
(
X
≤
93

18)
≤
279
279 + 18
2
=
.
4627
,
so
P
(
X >
75) = 1

P
(
X
≤
75)
≥
1

.
4627 =
.
5373. We conclude that
.
5373
≤
P
(
X >
75)
≤
1, so we have bounded
P
(
X >
75). (Note that the estimate from part (c),
P
(
X >
75)
≈
.