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MATH 264 PROBLEM HOMEWORK #1
Due to November 4, 2004@17:30
PROBLEMS
1. The numbers of full-length movies observed by a group of 400 persons were as follows:
Number of movies
0
1
2
3
4
5
6
7
8
Total
Number of persons
72
106
153
40
18
7
3
0
1
400
Find the median.
Solution. If we write the data as a list, we obtain that
0, 0, · · · , 0, 0, 1, 1, · · · , 1, 1, 2, 2, · · · , 2, 2, 3, 3, · · · , 3, 3, 4, 4, · · · , 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 8
|
{z
} |
{z
} |
{z
} |
{z
} |
{z
}
72 times
106 times
153 times
40 times
18 times
Since data is ordered from the smallest to the largest and the data size is even, the median is the
n+1
400 + 1
=
= 200.5 for the median position,
mean of the two middle number. Using formula
2
2
We see that the median is the mean of the 200th and 201st numbers, that is,
x̃ =
2+2
= 2.
2
2. For sample data {x1 , x2 , · · · , x25 }, it is given that
25
X
xi = 250,
and
25
X
i=1
x2i = 2884.
i=1
(a) Find x̄.
(b) Find s.
(c) Between what values, at least 75% of xi ’s, lie? (That is, you should find two numbers a and
b, such that at least 75% of xi ’s lie between a and b.)
Solution.
(a) Using the sample mean formula
Pn
x̄ =
we obtain that
x̄ =
i=1
n
xi
,
250
= 10.
25
(b) By shortcut formula for the sample standard deviation
s
Pn
Pn
2
2 − ( x=i xi )
x
i
i=1
n
s=
,
n−1
we obtain that
s
s=
2
2884 − (250)
25
=
24
r
2884 − 2500
=
24
r
384 √
= 16 = 4.
24
1
1
(or
1
−
100%) of the values lie
k2
k2
within k standard deviations on the either side of the mean. From the equality
1
1 − 2 100 = 75,
k
(c) According to the Chebyshev’s Theorem, at least 1 −
it follows that
1−
1
= 0.75
k2
⇒
1
= 0.25
k2
⇒
k2 =
1
0.25
⇒
k2 = 4
⇒
k = 2.
Thus, at least 75% of the values lie within 2 standard deviations on the either side of the mean.
In part (b), we have found that s = 4 and x̄ = 10, hence at least 75% of the values lie within
2 · 4 = 8 unit on the either side of 10. In other words, at least 75% of the values lie between
10 − 8 = 2 and 10 + 8 = 18.
3. (a) How many arrangements are there of all the letters in SOCIOLOGICAL?
(b) In how many of the arrangements in part (a) are A and G adjacent?
(c) In how many of the arrangements in part (a) are all the vowels adjacent?
Solution.
(a) Note that in SOCIOLOGICAL, S, G and A are seen once, C, I and L are seen twice, and O is
seen three times. Using the formula for the number distinguishable permutations of repeated
objects, we obtain that, the letters of SOCIOLOGICAL can be arranged in
12!
12 · 11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1
479001600
=
=
= 9979200
1!3!2!2!2!1!1!
1·3·2·1·2·1·2·1·2·1·1·1
48
ways.
(b) Let X denote the pattern AG. Then the permutations of letters S, O, C, I, O, L, O, X, I, C,
L corresponds to the permutations of the letters in SOCIOLOGICAL, in which A and G are
adjacent and A comes before G. The number of permutations of letters S, O, C, I, O, L, O, X,
I, C, L is equal to
11!
11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1
39916800
=
=
= 831600.
1!3!2!2!2!1!
1·3·2·1·2·1·2·1·2·1·1
48
Now let us denote by Y the pattern GA. Similarly, the permutations of letters S, O, C, I, O, L,
O, Y, I, C, L corresponds to the permutations of the letters in SOCIOLOGICAL, in which A
and G are adjacent and G comes before A, and this can also be done in 831600 ways. Hence,
totally in 831600 + 831600 = 1663200 of the arrangements of the letters in SOCIOLOGICAL,
A and G are adjacent.
(c) Now, we may denote by X = {X1 , X2 , · · · Xn } the set of permutations of the vowels O, O, O,
I, I, and A. It is easy to see that n (the total number of the permutations of the vowels in
SOCIOLOGICAL) is equal to
6!
720
=
= 60.
3!2!1!
12
Now, let Xi , i = 1, · · · 60, be an arbitrary permutation in X. Thus any permutation of the
letters S, C, C, L, L, G and Xi corresponds to a permutation of the letters in SOCIOLOGICAL
in which all the vowels are adjacent, and this can be done in
7!
5040
=
= 1260
1!2!2!1!1!
4
ways.
Since there are 60 Xi , and for each of them there are 1260 distinct permutations of the letters
in SOCIOLOGICAL, in which all the vowels are adjacent, totally in 60 · 1260 = 75600 of the
arrangements of the letters in SOCIOLOGICAL, all the vowels are adjacent.
4. (a) If two integers are selected, at random and without replacement, from {1, 2, 3, · · · , 99, 100},
what is the probability that the integers are consecutive?
2
(b) If two integers are selected, at random and without replacement, from {1, 2, 3, · · · , 99, 100},
what is the probability that their sum is even?
(c) If three integers are selected, at random and without replacement, from {1, 2, 3, · · · , 99, 100},
what is the probability that their sum is even?
Solution.
(a) Two integers are selected, at random and without replacement, from {1, 2, 3, · · · , 99, 100}, in
100
100 · 99
= 4950
|S| =
=
2·1
2
different ways.
Let A be the event that the selected integers are consecutive. Then A consists of pairs
{1, 2}, {2, 3}, {3, 4}, · · · , {99, 100} and so |A| = 99. Hence the desired probability is
P (A) =
|A|
99
1
=
=
= 0.02.
|S|
4950
50
(b) The sum of two integers is even if and only if: (i) both of the integers are even or (ii) both of
the integers are odd. Case (i) is possible in1
50
50 · 49
=
= 1225
2
2·1
ways. Similarly, Case (ii) is possible in 1225 ways. If B is the event that the sum of the
selected integers is even, then |B| = 1225 · 2 = 2450, and hence
P (B) =
2450
≈ 0.49.
4950
(c) Three integers are selected, at random and without replacement, from {1, 2, 3, · · · , 99, 100}, in
100
100 · 99 · 98
|S| =
=
= 161700
3
3·2·1
different ways.
The sum of three integers is even if and only if: (i) three of the integers are even, or (ii) two
of the integers are odd and one of the integers is even. Case (i) is possible in
50
50 · 49 · 48
=
= 19600
3
3·2·1
ways, and Case (ii) is possible in
50 50
50 · 49
=
· 50 = 61250
2
1
2·1
ways.
If C is the event that the sum of the selected integers is even, then |C| = 19600+61250 = 80850,
and hence
80850
P (C) =
= 0.5.
161700
1 Between
1 and 100, inclusive, there are 50 even numbers and 50 odd numbers.
3