Download Tutorial note of Math 144 8/Feb, 2010 1.A tire manufacturer wants to

Tutorial note of Math 144
8/Feb, 2010
1.A tire manufacturer wants to determine the inner diameter of
certain grade of tire. Ideally, the diameter would be 570 mm. The
data are as follows:
572 572 573 568 569 575 565 570 750
(a). Find the sample mean, median and mode?
(b). What feature in this data set is responsible for the substantial
difference between the sample mean and sample median?
(c). What can we conclude from last question?
(d). Find the sample variance, standard deviation and range.
Solution: (a). i. The sample mean is
x̄ =
572 + 572 + 573 + 568 + 569 + 575 + 565 + 570 + 750
= 590.44.
9
ii. For finding the median, we sort the observations in increasing
order:
565 568 569 570 572 572 573 575 750
Since n = 9, which is an odd number, thus the sample median is
= x5 = 572.
x̃ = x n+1
2
iii. Recall that the mode is the data that occurs most frequently,
so it is 572.
(b). 750 is an extreme observation.
1
(c). Mean is not robust, it is very sensitive to extreme values. In
this case it is suggested we use the median to measure the location.
(d). By definition of range,
range = max - min = 750 - 565 = 185,
the sample variance is
n
1 X
s =
(xi − x̄)2 = 3589.2,
n − 1 i=1
2
and the standard deviation is
v
u
n
u 1 X
t
s=
(xi − x̄)2 = 59.91.
n − 1 i=1
2. In a study of the relationship between heights and trunk diameters of trees, botany students collected sample data. Below are 20
tree circumferences (in feet):
1.8 1.9 1.8 2.4 3.1 3.4 3.7 3.7 3.8 3.9 4.1 4.9 5.5 8.3 13.7 4.0 5.1 5.1
5.2 5.3
(a). Find the five number summary and construct a box-plot.
(b). Identify the outlying values in the data.
Solution: (a). First sort the data in increasing order,
1.8 1.8 1.9 2.4 3.1 3.4 3.7 3.7 3.8 3.9 4.0 4.1 4.9 5.1 5.1 5.2 5.3 5.5
8.3 13.7
It’s easy to get min = 1.8; max = 13.7: To calculate lower quartile
Q1, noting that n = 20, p = 0.25, we get
np + 0.5 = 5.5, m = 5 < 5.5 < 6 = m + 1,
2
14
12
10
8
6
4
2
Figure 1: boxplot of question 2
Therefore the lower quartile is
1
1
Q1 = (xm + xm+1 ) = (x5 + x6 ) = (3.1 + 3.4)/22 = 3.25,
2
2
Similarly, we get the median
Q2 = (3.9 + 4.0)/2 = 3.95,
and the upper quartile
Q3 = (5.1 + 5.2)/2 = 5.15,
Hence, the boxplot is plotting in figure 1.
(b). We can get
Interquartile range (IQR) = Upper quartile - Lower quartile
= 5.15 - 3.25 = 1.9,
3
therefore,
Upper quartile + 1.5IQR = 5.15 + 2.85 = 8, and
Lower quartile - 1.5IQR = 3.25 - 2.85 = 0.4,
Since 8.3 > 8 and 13.7 > 8, then 8.3 and 13.7 are outliers.
3. The following data represent the length of life, in seconds, of 50
fruit flies subject to a new spray in a controlled laboratory experiment:
17 20 10 9 23 13 12 19 18 24 12 14 6 9 13 6 7 10 13 7 16 18 8 13 3
32 9 7 10 11 13 7 18 7 10 4 27 19 16 8 7 10 5 14 15 10 9 6 7 15
(a). Construct a double-stem-and-leaf plot for the life span of the
fruit ies using the stems 0?, 0•, 1?, 1•, 2?, 2•, and 3? such that
stems coded by the symbols ? and • are associated, respectively,
with leaves 0 through 4 and 5 through 9.
(b). Set up a relative frequency distribution.
Solution: (a). The double-stem-and-leaf plot is
0 ? |34
0 • |56667777777889999
1 ? |0000001223333344
1 • |5566788899
2 ? |034
2 • |7
3 ? |2
4
(b). The relative frequency distribution is given by
Interval Relative frequency
0-4
2/50
5-9
17/50
10-14
16/50
15-19
10/50
20-24
3/50
25-29
1/50
30-34
1/50
5