Assignment # 2 Solution – Statistics

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Assignment # 2 Solution - Statistics

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Assignment from Chapter 2: solution
2.1( b) The mean is the sum of the measurements divided by the number of measurements, or
0 5 1 1 3 10
2
5 5
i x
x
n
= Σ = + + + + = =
To calculate the median, the observations are first ranked from smallest to largest: 0, 1, 1, 3, 5. Then
since n = 5 , the position of the median is 0.5(n +1) = 3 , and the median is the 3rd ranked
measurement, or m =1 . The mode is the measurement occurring most frequently, or mode = 1.
2.3 a
58
5.8
10
i x
x
n
Σ = = =
b The ranked observations are: 2, 3, 4, 5, 5, 6, 6, 8, 9, 10. Since n = 10 , the median is halfway
between the 5th and 6th ordered observations, or m = (5 + 6) 2 = 5.5 .
c There are two measurements, 5 and 6, which both occur twice. Since this is the highest frequency of
occurrence for the data set, we say that the set is bimodal with modes at 5 and 6.
2.8 a Similar to previous exercises. The mean is
0.99 1.92 0.66 12.55
0.896
14 14
i x
x
n
= Σ = + + + = = L
b To calculate the median, rank the observations from smallest to largest. The position of the median
is 0.5(n +1) = 7.5 , and the median is the average of the 7th and 8th ranked measurement or
m = (0.67 + 0.69) 2 = 0.68 .
c Since the mean is slightly larger than the median, the distribution is slightly skewed to the right.
2.10 a Similar to previous exercises.
2150
215
10
i x
x
n
Σ = = =
b The ranked observations are shown below:
175 225
185 230
190 240
190 250
200 265
The position of the median is 0.5(n +1) = 5.5 and the median is the average of the 5th and 6th
observation or
200 225
212.5
2
+ =
c Since there are no unusually large or small observations to affect the value of the mean, we would
probably report the mean or average time on task.
2.14 a
12
2.4
5
i x
x
n
Σ = = =
b Create a table of differences, (xi – x ) and their squares, ( )2
i x – x .
i x i x – x ( )2
i x – x
2 –0.4 0.16
1 –1.4 1.96
1 –1.4 1.96
3 0.6 0.36
5 2.6 6.76
Total 0 11.20
Then
( )2 2 2
2 (2 2.4) (5 2.4) 11.20
2.8
1 4 4
i x x
s
n
Σ – = = – + + – = =
–
L
c The sample standard deviation is the positive square root of the variance or
s = s2 = 2.8 = 1.673
d Calculate 2 22 12 52 40 i Σ x = + +L+ = . Then
( ) ( ) 2 2
2
2
12
40 11.2 5 2.8
1 4 4
i
i
x
x
s n
n
Σ
Σ – –
= = = =
–
and s = s2 = 2.8 = 1.673 .
The results of parts b and c are identical.
2.15 a The range is R = 4 -1 = 3 .
b
17
2.125
8
i x
x
n
Σ = = =
c Calculate 2 42 12 22 45 i Σ x = + +L+ = . Then
( ) ( ) 2 2
2
2
17
45 8.875 8 1.2679
1 7 7
i
i
x
x
s n
n
Σ
Σ – –
= = = =
–
and s = s2 = 1.2679 = 1.126 .
2.17 a The range is R = 2.39 -1.28 = 1.11 .
b Calculate 2 1.282 2.392 1.512 15.415 i Σ x = + +L+ = . Then
( ) ( ) 2 2
2
2
8.56
15.451 .76028 5 .19007
1 4 4
i
i
x
x
s n
n
Σ
Σ – –
= = = =
–
and s = s2 = .19007 = .436
c The range, R = 1.11 , is 1.11 .436 = 2.5 standard deviations.
2.19 a The range of the data is R = 6 -1 = 5 and the range approximation with n = 10 is 1.67
3
R
s » =
b The standard deviation of the sample is
( ) ( ) 2 2
2
2
32
130
10 3.0667 1.751
1 9
i
i
x
x
s s n
n
Σ
Σ – –
= = = = =
–
which is very close to the estimate for part a.
2.20 a First calculate the intervals:
36 3 or 33 to 39
2 36 6 or 30 to 42
3 36 9 or 27 to 45
x s
x s
x s
± = ±
± = ±
± = ±
According to the Empirical Rule, approximately 68% of the measurements will fall in the interval 33
to 39; approximately 95% of the measurements will fall between 30 and 42; approximately 99.7% of
the measurements will fall between 27 and 45.
b If no prior information as to the shape of the distribution is available, we use Tchebysheff’s
Theorem. We would expect at least (1-1 12 ) = 0 of the measurements to fall in the interval 33 to
39; at least (1-1 22 ) = 3 4 of the measurements to fall in the interval 30 to 42; at least
(1-1 32 ) = 8 9 of the measurements to fall in the interval 27 to 45.
2.27 a The centre of the distribution should be approximately halfway between 0 and 9 or (0 + 9) 2 = 4.5 .
b The range of the data is R = 9 – 0 = 9 . Using the range approximation, s » R 4 = 9 4 = 2.25 .
c Using the data entry method the students should find x = 4.586 and s = 2.892 , which are fairly
close to our approximations.
2.31 a We choose to use 12 classes of length 1.0. The tally and the relative frequency histogram follow.
Class i Class Boundaries Tally fi Relative frequency, fi/n
1 2 to < 3 1 1 1/70
2 3 to < 4 1 1 1/70
3 4 to < 5 111 3 3/70
4 5 to < 6 11111 5 5/70
5 6 to < 7 11111 5 5/70
6 7 to < 8 11111 11111 11 12 12/70
7 8 to < 9 11111 11111 11111 111 18 18/70
8 9 to < 10 11111 11111 11111 15 15/70
9 10 to < 11 11111 1 6 6/70
10 11 to < 12 111 3 3/70
11 12 to < 13 0 0
12 13 to < 14 1 1 1/70
TREES
Relative frequency
5 10 15
20/70
10/70
0
b Calculate 70, 541 i n = Σ x = and 2 4453 i Σ x = . Then
541
7.729
70
i x
x
n
Σ = = = is an estimate of m .
c The sample standard deviation is
( ) ( ) 2 2
2 541
4453
70 3.9398 1.985
1 69
i
i
x
x
s n
n
Σ
Σ – –
= = = =
–
The three intervals, x ± ks for k =1,2,3 are calculated below. The table shows the actual
percentage of measurements falling in a particular interval as well as the percentage predicted by
Tchebysheff’s Theorem and the Empirical Rule. Note that the Empirical Rule should be fairly
accurate, as indicated by the mound-shape of the histogram in part a.
2.63 a Answers will vary. A typical histogram is shown below. The distribution is slightly skewed to the left.
8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Fuel Efficiency
Relative frequency
b Calculate 20, 193.1 i n = Σ x = , 2 1876.65 i Σ x = . Then
i 9.655 x
x
n
Σ = =
( ) ( ) 2 2
2 193.1
1876.65
20 .646 0.804
1 19
i
i
x
x
s n
n
Σ
Σ – –
= = = =
–
c The sorted data is shown below:
7.9 8.1 8.9 8.9 9.4 9.4 9.5 9.5 9.6 9.7
9.8 9.8 9.9 9.9 10.0 10.1 10.2 10.3 10.9 11.3
The z-scores for x = 7.9 and x = 11.3 are
7.9 9.655 11.3 9.655
2.18 and 2.05
.804 .804
x x x x
z z
s s
= – = – = – = – = – =
Since neither of the z-scores are greater than 3 in absolute value, the measurements are not judged
to be outliers.
d The position of the median is 0.5(n +1) = 10.5 and the median is m = (9.7 + 9.8)/2 = 9.75
e The positions of the quartiles are 0.25(n +1) = 5.25 and 0.75(n +1) = 15.75 . Then
1 Q = 9.4 + 0.25(9.4 – 9.4) = 9.4 and 3 Q = 10.0 + 0.75(10.1-10.0) = 10.075.
k x ± ks Interval Fraction in Interval Tchebysheff Empirical Rule
1 7.729 ±1.985 5.744 to 9.714 50/70 = 0.71 at least 0 » 0.68
2 7.729 ± 3.970 3.759 to 11.699 67/70 = 0.96 at least 0.75 » 0.95
3 7.729 ± 5.955 1.774 to 13.684 70/70 = 1.00 at least 0.89 » 0.997