Download Solution to quiz 1

Solution to ST361 Quiz 1, 1/27/2009
1. The following histogram (using the convention of ”[, )”) is for the carbon
monoxide levels (in ppm) in air samples obtained from a certain region:
Histogram of monoxide levels
6
5
Frequency
4
3
2
1
0
6
8
10
12
14
16
18
Monoxided levels in ppm
1a (10pts) What is the shape of the distribution of the data, symmetric,
positively skewed or negatively skewed?
Positively skewed (or skewed to right)
1b (10 pts) Which of the following relationship between the sample mean and
sample median is most likely to be true?
(i) mean = median (ii) mean > median
(iii) mean < median
(ii) is the most likely to be true
1c (10pts) The y-axis gives the frequency. Can you tell how many data points
are used to construct the histogram? How many carbon monoxide levels are
less than 10?
The total number used is : 1+6+4+2+1+1 = 15
The number of carbon monoxide levels that are less than 10 is 1+6=7.
1d (10pts) Convert the histogram using relative frequency (you can just mark
the relative frequency in the previous histogram). Find the proportion that the
carbon monoxide levels are less than 12.
1
Class
Frequency
Relative frequency
[6,8)
1
6.7%
[8,10)
6
40%
[10,12)
4
26.7%
[12,14)
2
13.3%
[14,16)
1
6.7%
[16,18)
1
6.7%
The proportion that the carbon monoxide levels are less than 12 is
6.7+40+26.7 = 73.4 (%)
2. The following data shows the amount of aluminum contamination in plastic
of certain type: 60, 30, 125, 63, 70,120, 79, 102, 90, 86, 115, 101, 119. Do
the following:
2a (10 pts) Find the sample mean and sample median.
Mean = 89.2
The sorted data is: 30 60 63 70 79 86 90 101 102 115 119 120 125, so
Median = 90
2b (10pts) Find the sample variance and sample standard deviation (you are
given
x
s2 
2
i
 113122).
1
1
  xi2  n( x ) 2  
113122 - 13  89.22   807.14
n 1
13  1
s  s 2  807.14  28.4
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2c (10pts) Construct a stem-and-leaf plot for the data.
3|0
4|
5|
6 | 03
7 | 09
8|6
9|0
10 | 12
11 | 59
12 | 05
2d (10pts) Find the first quartile and the third quartile and the interquartile
range (IQR).
Q1=70, Q3=115, IQR=115-70=45
2e (10 pts) Are there any outliers in the data set?
Q1 – 1.5IQR = 2.5 (no data point smaller than 2.5)
Q3 + 1.5IQR = 182.5 (no data point larger than 182.5)
No outlier
2f (10pts) Construct a boxplot for the data.
3