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SOLUTION to MATH 264 HOMEWORK 1
1) Find the mean, the median, the mode, the range, the variance, the standard deviation, the quartiles
Q1 and Q3 of the following sample:
41
35
19
18
44
39
13
49
11
16
42
17
Solution: To find the sample mean we calculate
P20
i=1
x̄ =
20
26
38
14
P20
i=1
xi
20
=
47
12
13
26
xi . It is 540. Therefore, the sample mean is
540
= 27
20
Since 13 and 26 are the most frequent numbers. (repeated 2 times) the sample modes are 13 and 26
To find the sample median we need to order the elements from smallest to largest (or largest to smallest)
11
26
12
26
The median position is found from
and the eleventh elements, namely
13
35
13
38
14
39
16
41
17
42
18
44
19
47
20
49
20 + 1
= 10.5. So the sample median x̃, is the mean of the tenth
2
x̃ =
The position of first quartile Q1 is found from
elements, namely Q1 = 14.5
20 + 26
= 23.
2
20 + 1
= 5.25. So Q1 is between the fifth and the sixth
4
3(20 + 1)
The position of third quartile Q3 is found from
= 15.75. So Q3 is between the fifteenth and
4
the sixteenth elements, namely Q3 = 40.5
The sample range is the largest value minus the smallest, namely, 49 − 11 = 38.
To evaluated the sample variance s2 ,we will use the computational formula:
P
P20 2 ( 20
xi )2
i=1
x
−
i=1 i
2
20
s =
.
19
P20
First, we compute i=1 x2i . It is 17962. Then the sample variance is
2
17962 − (540)
20
s =
= 178.
19
2
√
The standard deviation is s = s2 ≈ 13.34
2) A travel brochure lists 9 museums in the city of İstanbul. A tourist is planning to spend 3 days in
Istanbul and visit 2 museums each day. In how many ways can he visit museums if the order of the
museums he visited during each day, is important?
Solution: In
9 · 8 · 7 · 6 · 5 · 4 = 60480
ways.
3) Consider the following two sets of data
V alue
F requency
2 3
5 3
4
10
5
10
6
10
7
10
8
5
9
1
10
5
11
1
V alue
F requency
2 3
1 1
4
10
5
15
6
13
7
13
8
5
9
1
10
0
11
1
Construct frequency distribution tables and frequency histograms for these two sets of data.
Solution: Step 1: There are 60 values in both data sets, so we need 6 frequency classes for both data
sets.
Step 2: The smallest value is 2 and the largest value is 11 for both data sets. So the range (of data)
is 11-2=9.
Step 3: The data is given in integers (whole numbers), so the class range is an integer. It is 9/6=1.5
rounded up to 2.
Frequency Distribution Table for the first set of data is
Class
Class 1
Class 2
Class 3
Class 4
Class 5
Class 6
Boundaries
1.5-3.5
3.5-5.5
5.5-7.5
7.5-9.5
9.5-11.5
11.5-13.5
Total
Frequency
8
20
20
6
6
0
60
Frequency Distribution Table for the second data set is
Class
Class 1
Class 2
Class 3
Class 4
Class 5
Class 6
Boundaries
1.5-3.5
3.5-5.5
5.5-7.5
7.5-9.5
9.5-11.5
11.5-13.5
Total
Frequency
2
25
26
6
1
0
60
The histogram for the first data set can be drawn as:
F 2020R
E
Q
U
E
N 8C 6Y
|
2
|
4
|
6
|
8
|
10
Similarly, the histogram corresponding to the second data set can be constructed.
4) A sample of 450 workers received an average weekly wage of $ 500 and the variance of the wages is
$2500.
a) At most how many workers will earn less that $ 350 or more that $650?
b) At least how many workers will earn between $ 350 and $ 650 a week?
Solution: (b) Since standard deviation of an average weekly wage is s = $50, then interval between $350
and
$650
¡
¢ is within 3 standard deviations from the mean x̄ = $500. By Chebyshev’s Theorem, at least
1 − 312 × 450 = 400 workers will earn between $ 350 and $ 650 a week.
(a) It follows from part (b) that less that 450-400=50 workers will earn less that $ 350 or more that $650.
5) A poker hand contains five cards dealt from a deck of 52. How many distinct poker hands can be
dealt containing:
1. two pairs (for example, 2 kings, 2 aces, and a 3)?
2. a flush (five cards in a given suit)?
3. a straight flush (any five in sequences in a given suit, but not including ten, jack, queen, king, ace)?
4. a royal flush (ten, jack, queen, king, ace in a single suit)?
¡ ¢
Solution: If there is no condition totally 52
5 distinct poker hands are possible.
¡ ¢
1. Since there are 13 denominations, for the denominations involved in the two pairs, there are 13
2
possibilities. Since there ¡are
4 cards of a given denomination, the number of possibilities for a pair in
¢
a given denomination is 42 . Thus the number of possibilities for selecting four cards such that two
¡ ¢ ¡4¢ ¡4¢
pairs of different denominations are involved is 13
2 · 2 · 2 . For the fifth card, 11 denominations
with four cards in each denomination are possible. By the basic counting principle, the number of
two-pair hands that are possible is
µ ¶ µ ¶ µ ¶
13
4
4
13 · 12 4 · 3 4 · 3
·
·
· 44 =
·
·
· 44 = 123, 552.
2
2
2
2·1 2·1 2·1
¡¢
2. Since there are 4 suits, we can choose the suit involved in the flush in 41 = 4 possible ways. Since
¡ ¢
there are 13 cards of a given suit, five cards from a given suit can be chosen in 13
5 possible ways.
Thus the number of flush hands that are possible is
µ ¶
13
13 · 12 · 11 · 10 · 9
4·
=4·
= 5, 148.
5
5·4·3·2·1
3. Since there are 4 suits, we can choose the suit involved in the straight flush in
ways. Ordered fives in a given suit may be one of
¡4¢
1
= 4 possible
{A-2-3-4-5; 2-3-4-5-6; · · · ; 8-9-10-J-Q; 9-10-J-Q-K}.
Thus, the number of straight flush hands that are possible is
4.9 = 36
(or, 4.8 = 32 if you exclude the sequence A-2-3-4-5).
4. Now the answer is 4, because after you choose the suit, there is only one way of having a royal flush
hand, namely 10-J-Q-K-A.
6) A class has 10 students. A history instructor wants to split these 10 students into 3 discussion groups.
One group will consist of 4 students and discuss topic A. The second and third groups will discuss topics
B and C respectively, and consist of 3 students each. In how many ways can the instructor form the
groups?
Solution: In
µ ¶ µ ¶ µ ¶
10
6
3
10 · 9 · 8 · 7 6 · 5 · 4
·
·
=
·
· 1 = 4, 200
4
3
3
4·3·2·1 3·2·1
ways.
7) Compute mean,
P
Pn variance and standard deviation for a sample consisting of n = 10 items, and having
n
2
x
=
84,
i=1 i
i=1 xi = 20.
Solution: The sample mean is
Pn
i=1
x̄ =
xi
10
=
20
= 2.
10
The sample variance is
P10
2
s =
2
i=1 xi −
9
(
P10
i=1
xi ) 2
10
The sample standard deviation is
s=
√
2
84 − 20
44
10
=
=
≈ 4.89.
9
9
s2 ≈ 2.21.