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Unformatted text preview: Student: Grady Simonton
Instructor: Shawn Parvini Date: 2/18/10 Course: Math119: Elementary Statistics  Spring 2010  CRN: 49239  16 weeks
Book: Triola: Elementary Statistics, 11e Time: 11:01 ANI The data represents the daily rainfall (in inches) for one month. {143 0 0 023 0 (141 Construct a frequency distribution beginning with a lower class 0 0.23 0 0 1.32 0 limit of 0.00 and use a class width of 0.20. Does the frequency distribution appear to be roughly a normal distribution? 0‘ 19 0 0‘02 0 0‘2 l 0
0.11 0.62 0 0.02 0 0.26
0 0.19 0 0 0.21 0 First decide on the number of classes you want. The formula for fmding the class width is given below. . (maximum value)  (minimum value)
Class Width =2 —
number of classes Since the class width, maximum value, and minimum value are given, determine the number of classes. 0 20 1.32  0
. ‘3
number of classes number of classes as 6.6
as 7" Use the lower limit of the ﬁrst class and the class width to ﬁll in the ﬁrst class. Daily Rainfall
(in inches)
0000. 19 ? Frequency Using the lower limit of the ﬁrst class and the class width, proceed to list the other lower class limits. (Add the class width to the starting point to get the second lower class limit. Add the class width to the second lower class limit to get
the third, and so on.) Now ﬁll in the next class. Daily Rainfall Frequency
(in inches)
0.000.19 ?
0.200.39 ? The remaining ﬁve classes are given below. Daily_ Rainfall Frequency Daily. Rainfall Frequency
(m inches) (in mches)
0.000.19 ? 0.800.99 ?
0.200.39 ? 1.001.19 ?
0.400.59 ? 1.201.39 ?
0.60—0.79 ? Page 1 Student: Grady Simonton Course: Math119: Elementary Statistics  Spring 2.010  CRN: 492.39
Instructor: Shawn Parvini  16 weeks Date: 2/18/10 Book: Triola: Elementary Statistics, 11e
Time: 11:01 ANT To determine the frequencies, go through the data set putting a tally in the appropriate class for each data value. Use the
tally marks to find the total frequency for each class. Start with the ﬁrst frequency. There are 2] rainfall amounts between 0.00 and 0.19. There are 16 days with no rain, 2
days with 0.02 inches of rain, 1 day with 0.11 inches of rain, and 2 days with 0.19 inches of rain. Daily. Rainfall Frequency DailyRainfall Frequency
(1n mches) (m 1nches)
0.000.19 21 0.800.99 ‘?
0.200.39 ? 1.001.19 ?
0.400.59 ? 1.201.39 ?
0.600.79 ? Now determine how many rainfall amounts are between 0.20 and 0.39. Daily Rainfall Frequency Daily Rainfall Frequency
(in inches) (in inches)
0.000.19 21 0.800.99 ?
0.200.39 5 1.001.19 ‘?
0.400.59 ? 1.201.39 ?
0.600.79 ? Use the same procedure to ﬁll in the remaining frequencies. Daily Rainfall Frequency Daily. Rainfall Frequency
(in inches) (in inches)
0.000.19 21 0.800.99 0
0.200.39 5 1.001.19 0
0.400.59 2 1.201.39 l
0.600.79 1 For now, we can judge that a frequency distribution is approximately normal by determining whether it has the
following features. 1. The frequencies start low, then increase to some maximum frequency, then decrease to a low frequency.
2. The distribution should be approximately symmetric, with frequencies evenly distributed on both sides of the
maximum frequency. The frequencies do not start low, then increase to some maximum frequency, then decrease to a low frequency. Therefore, the distribution does not appear to be normal. Page 2. ...
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 PARVINI

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