1. Week 2 LT Assignment DUE: 10/17/2011 MTH233
Team B: Katie Johnson, Laura Laskey, Rosana Cedeno, & Taci Toves

2. Create a frequency table with these data. (Date sheet)
The ages (years) for a random sample of 35 participants in the 2007 Youngstown Peace Race 12 14 15 17 19 20 23 24 26 27 33 37 39 40 41 44 45 47 51 52 56 59
1146
Total age of Participants
Median
Katie Johnson

3. Create a frequency table with these data.
Bin
Frequency 14 2 23 11 32 3 41 7 50 59 5
Frequency Table
Age of Participants
Number of Participants
6 – 14 2
15 – 23 11
24 – 32 3
33 – 41 7
42 – 50
51- 59 5
MODE is 11 participants in age group 2 ,15 to 23 years olds.  We also have to age groups that have the same number of participants, 7 in age group 4 & 5.
Katie Johnson

4. Construct a histogram of these data, and describe its shape.
The histogram suggests there are no presents of outliers. I believe that the shape is bimodal. It has two peaks with age group 2 and age groups 4 & 5.
According to our class lecture (MATH 233-Week 2 Lecture 2011) quartiles (Q1 & Q3) can be used to determine the Five Number Summary and Turkey’s Rule uses the Interquartile Range (IQR to determine outliers. Therefore, using Example 5 from our lecture for Q1 & Q3.
Q1 = (35)(0.25)
Q1 = 8.25 rounding up to next whole number is 9
Q1-9
Q3 = (35)(0.75)
Q3 = rounding up to 27
Q3 = 27
Using Example 6 from our lecture for finding the Five Number Summary for the ages of the sampled participants are as follows:
Min: 14
Q1: 17
Median: 37 (taken from Excel sheet using the median function.)
Q3: 44
Max: 59
Then using the Interquartile Range (IQR) and Tukey’s Rule we can determine the outliers.
IQR = Q3-Q1
IQR = 44-17
IQR = 27
Therefore Tukey’s Rule is calculated as follows:
IQR x 1.5 = T
27 x 1.5 = T
40.5 = T
Q1 – T = low outliers
= low outliers
-33.5 = low outliers
Q3 + T = high outliers
= high outliers
84.5 = high outliers
Katie Johnson

5. Find the mean, median, and standard deviation of the data values.52 44 20 51 56 14 39 37 17 15 40 23 26 24 45 47 19 12 27 59 41 33
STDEV
MEAN
MEDIAN
14.4
32.7
The mean(32.7) and median(37) compared to the mode(11) indicate that the graph should be skewed to the right or Positively Skewed.
Taci Toves

6. Does the empirical rule hold for these data?
The Empirical Rule does hold true for the given data. As discussed in Week 2’s Lecture (p. 7), the empirical rule shows that the data follows a normal distribution by relating the mean and standard deviation. See the attached document (Excel spreadsheet- mean and standard deviation) in which I calculated the mean of the data to be and the standard deviation The Empirical Rule states that 68% of the data will fall within one standard deviation of the mean, 95% of the data will fall within two standard deviations of the mean, and 99.7% of the data will fall within three standard deviations of the mean (Week 2 Lecture, pp. 7. 8).
For the given data, one standard deviation from the mean would be / The data needs to fall between and
Two standard deviations from the mean would be / (which is 2 times the standard deviation of ). The data needs to fall between 3.9 and
Three standard deviations from the mean would be / (which is 3 times the standard deviation of ). The data needs to fall between and 76.
My calculations show that 21 of the 35 ages fall within one standard deviation from the mean, which calculates to 60%. (68% was predicted by the Empirical Rule)
35 of the 35 ages in the data fall within two standard deviations from the mean, which calculates to 100%. (95% was predicted by the Empirical Rule)
35 of the 35 ages in the data fall within three standard deviations from the mean, which calculates to 100%. (99.7% was predicted by the Empirical Rule)
Therefore, the observed data results are consistent with a normal distribution. See the attached table (Next Slide) to compare the observed and predicted percentages for the data.  
Laura Laskey

7. Does the empirical rule hold for these data? (Cont’d-Empirical Table)
Interval
Percent predicted by empirical rule
Percent observed
13.32 to 47.16
68%
60%
3.9 to 61.58
95%
100%
to 76
99.70%
Empirical Rule- Predicted and Observed percentages
Laura Laskey

8. Do the computations for the previous question support your claim about the presence of outliers? Explain. 52 44 20 51 56 14 39 37 17 15 40 23 26 24 45 47 19 12 27 59 41 33
These computations support the claim that there are no outliers in the data, because 100% of the data falls within two standard deviations of the mean. If outliers were present, some of the data would fall outside of three standard deviations of the mean. That is not the case with the given data, so there are no outliers.
Laura Laskey

9. Write a short summary of your findings about the distribution of ages.
Rosana Cedeno