.
CLABE Statistics
Homework assignment - Problem sheet 6
.
1.
The students at a local university spend a lot of money each term on textbooks. Suppose that
the amount of money spent on textbooks for a term, denoted by
with a mean of
e160
and a standard deviation of
X
e20.
in the panel below;
f(x)
(a) Sketch the density function of
X , follows a normal distribution
x
(b) the local bookstore will oer a t-shirt to any student who spends more than
e200
on
textbooks. What proportion of students are eligible for the t-shirt?
(c) Compute the probability that a student selected at random will spend between 100 and 150
euros on textbooks;
(d) Justin spent
e170
on textbooks. What percentile does this value of
e170
correspond to?
(e) Complete the sentence: Based on the above model, 20% of the students spend less than
e____
on textbooks.
SOLUTION
In the following,
Z
will always denote a random variable following a standard normal distribution.
(a) The the plot of the density function of
X
1
is
0.02
0.01
0.00
f(x)
80
100
120
140
160
180
200
220
240
x
(b) The proportion of students eligible for the t-shirt is equal to the probability that a randomly
selected student will spend more than 200 euros on textbooks. This amounts to 2.28% as
computed below,
P (X > 200) = 1 − P (X ≤ 200) = 1 − P (Z ≤ 2) = 1 − Φ(2) = 1 − 0.9772 = 0.0228.
(c)
P (100 < X < 150) = P (−3 < Z < −0.5) = Φ(−.5) − Φ(−3) = 0.3085 − 0.0013 = 0.3072.
(d) This percentile is the proportion of students who spend less than
e170
on textbooks, that
is
P (X ≤ 170) = P (Z ≤ 0.5) = Φ(0.5) = 0.6915.
(e) Since the equation
P (X < x) = 0.20 is satised for x = 160−0.84×20 = 143.2, the sentence
can be completed as follows:
143.2 on textbooks.
e
Based on the above model, 20% of the students spend less than
2.
Half Moon Bay, California, has an annual pumpkin festival at Halloween. A prime attraction to
this festival is a largest pumpkin contest. Suppose that the weights of these giant pumpkins
are approximately normally distributed with a mean of 125 pounds and a standard deviation of
18 pounds. Farmer Harv brings a pumpkin that is at the 85% percentile of all the pumpkins in
the contest. What is the approximate weight of Harv's pumpkin?
2
SOLUTION
If we denote by
X
the weight of the pumpkins, then the approximate weight of Harv's pumpkin
x
is given by the value
z = 1.03
satised for
P (X ≤ x) = 0.85. Since the
x = 125 + 1.03 × 18 = 143.54 pounds.
such that
then
equation
P (Z ≤ z) = 0.85
is
3.
A random variable
X ∼ N (µ, σ 2 ).
X
is normally distributed with mean
µ,
and standard deviation
σ;
that is,
What is the probability that a term selected at random from this population will
be more than 2.5 standard deviations from the mean?
SOLUTION
P ({X < µ − 2.5 × σ} ∪ {X > µ + 2.5 × σ}) = 1 − P (µ − 2.5 × σ < X < µ + 2.5 × σ)
= 1 − P (−2.5 < Z < 2.5)
= 1 − [Φ(2.5) − Φ(−2.5)]
= 1 − [0.9938 − 0.0062] = 0.0124.
4.
The normal random variable
0.90.
Find the mean
µ
X
has a standard deviation of 12. We also know that
P (X > 50) =
of the distribution.
SOLUTION
Since
P (Z > −1.28) = 0.90
(50 − µ)/12 = −1.28
then
and therefore
µ = 50 + 1.28 × 12 = 65.36.
5.
You took a multiple-choice test.
Suppose that the average test score is 70 with a standard
deviation of 5 and you scored 82. If the distribution of scores is approximately normal, then how
good did you do?
SOLUTION
If
X
denotes the score of a student selected at random, then
P (X ≤ 82) = P (Z ≤ 2.4) = 0.9918
and therefore you did very well because your score corresponds to the 99.18 percentile of all the
scores.
6.
Let
X1
and
X2
(a) Calculate
two i.i.d. random variables with discrete uniform distribution on the set
E(X1 ),
Var(X1 ),
E(X2 )
and Var(X2 ).
(b) Find the probability mass function of
(c) Calculate
E(S)
S = X1 + X2 .
and Var(S).
3
{0, 1, 2}.
D = X1 − X2 .
(d) Find the probability mass function of
(e) Calculate
E(D)
and Var(D).
(f ) Verify that
E(D) = E(X1 ) − E(X2 )
E(S) = E(X1 ) + E(X2 ),
and that
Var(S)
=Var(D) =Var(X1 )+Var(X2 ).
SOLUTION
(a) Since
X1
and
X2
E(X1 ) = E(X2 ) and Var(X1 )=Var(X2 ).
E(X1 ) = 1, E(X12 ) = 35 and Var(X1 ) = 35 − 1 = 23 .
are identically distributed, then
Furthermore, it is easy to compute
(b) The possible values of
s = x1 + x2
for
x1 = 0, 1, 2
x2 = 0, 1, 2
and
are as follows
X1
X2
+
0
1
2
0
0
1
2
1
1
2
3
2
2
3
4
Every entry of the above table has probability
S
1/9
so that the probability mass function of
is
s
0
1
2
3
4
P (S = s)
1
9
2
9
3
9
2
9
1
9
S
−4=
(c) One can use the probability distribution of
E(S) = 2, E(S 2 ) =
48
9 and Var(S)
(d) The possible values of
d = x1 − x2
=
48
9
for
computed in the previous point to see that
4
3.
x1 = 0, 1, 2
and
x2 = 0, 1, 2
are as follows
X1
X2
−
0
1
0
0
1
2
1
-1
0
1
2
-2
-1
0
Every entry of the above table has probability
D
2
1/9
so that the probability mass function of
is
d
-2
-1
0
1
2
P (D = d)
1
9
2
9
3
9
2
9
1
9
(e) One can use the probability distribution of
E(D) = 0,
Var(D)
=
E(D2 )
=
4
3.
4
D
computed in the previous point to see that
(f ) This check can be easily carried out by comparing the results of point (a) above with the
results of points (c) and (e).
7.
Let
the
X1 , X2 , . . . , X50
set {0, 1, 2}.
a sequence of i.i.d. random variables with discrete uniform distribution on
(a) Find the support of the random variable
(b) Calculate
E(S)
S
obtained as
S = X1 + X2 + · · · + X50 ;
and Var(S);
(c) nd the approximate distribution of
S;
(d) nd the support of the random variable
X̄ =
(e) nd the approximate distribution of
X̄
given by
X1 + X2 + · · · + X50
;
50
X̄ ;
(f ) calculate the approximate probability that
35 ≤ S ≤ 65;
(g) compute the approximate probability that
0.7 ≤ X̄ ≤ 1.3.
SOLUTION
(a) The random variable
S
can take on any value in the set
(b) As shown in the previous exercise
a
X1 , X2 , . . . , X50
are i.i.d., then
{0, 1, 2, . . . , 98, 99, 100}
E(Xi ) = 1 and Var(Xi ) = 23 for every i = 1, . . . , 50.
E(S) = 50 and Var(S) = 100
3 .
(c) It is possible to apply the central limit theorem to show that
(d) The random variable
(e)
X̄
S ≈ N (50, 100/3).
can take on any value in the set
0 1 2
98 99 100
,
,
,..., ,
,
50 50 50
50 50 50
E(X̄) = E(Xi ) = 1
Since
and Var(X̄)
theorem the distribution of
X̄
that is
{0, 0.02, 0.04, . . . , 1.96, 1.98, 2}
=Var(Xi )/50 = 2/150.
Furthermore, by the central limit
is approximatively normal, that is
X̄ ≈ N (1, 2/150).
(f )
P (35 ≤ S ≤ 65) ≈ P (−2.59 ≤ Z ≤ 2.59) = Φ(2.59) − Φ(−2.59) = 0.9952 − 0.0048 = 0.9904.
(g) This probability can be computed by noticing that
S = 50 × X̄ and that 35 = 50 × 0.7
1.3) = P (35 ≤ S ≤ 65) = 0.9904.
also note that
5
X̄ ≈ N (1, 2/150).
65 = 50 × 1.3 so
and
However, one can
that
P (0.7 ≤ X̄ ≤