Week 6, Lab 7
For this deliverable, you will work on an
Excel but you should post your answers here
Due before class on Monday 10/5 (the class following next class)
OBS! Like with the previous deliverable, you are expected to work in Excel to answer all the questions, but you do not need to upload your Excel document. You should post all answers as a response to this post.
Instructions
I. Download the database named “Lab_6_the-counted-2016_simplified.csv”. You have already used this simplified version of the database “The Counted”. Next, you should open and compute the data in Excel, but answer the questions in this post.
II. Download the instructions and follow them in order to complete the computation. They are called “Lab7 Measures of dispersion.docx”.
III. After computing all requested tasks in Excel, please answer the questions below.
Questions (answer all of them and consult Lab 7 video to complete the answers)
1. Are there many modes for the age variable? Which ones are they/is it? What does it mean?
2. Is the age variable skewed? If so, how and why?
3. What are the minimum, maximum and range values of age?
4. What is the interquartile range? What does this mean? Hint: watch the video to inform your explanation.
5.) What is the standard deviation of the age variable? How do you interpret the standard deviation?
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1. 25.
2. It is not skewed because of the closeness in range.
3. Minimum = 3
Maximum = 87
Range = 84
4.
•Interquartile range: 18.
• Q1 and Q3 numbers are 27 and 45, which means that all the numbers before 27 and after 45 are outliers.
5. The standard deviation is 13.14 and it is closer to the mean.
1. The mode for the age variable is 25
2. The average age is 36.8 and the median age 35. Resulting in the age variable not being skewed,
3.The minimum is, max is 87 & range of 84
4. The Interquatile Range is 18
5. The standard deviation is 13.13
1.The mode is 25
2. Its not skewed because the mean and median are close.
3. The minimum is 3. The maximum is 87. And the range is 84
4. The interquartile is 18
5. The standard deviation is 13.13
1. The mode for the age variable is 25
2. Average age (36.8) Median age (35) Since those numbers are close the age variable is not skewed
3.Min: 3 Max: 87 Range: 84
4. The Interquatile Range is 18 which is in between Q1 and Q3 which leaves 47 being the 75th percentile and 27 being the 25th percentile as the outliers.
5. The standard deviation is 13.13 which means the data for the age variables has less diversity.
1. The mode is 25
2. The mean is 36.81 and the median is 35 therefor the age variable is not skewed.
3. Minimum=3
Maximum= 87
Range=84
4. The interquartile range is 18. That let us know that the numbers before 27 and after 45 are outliers.
5. The standard deviation is 13.14, meaning they are close to the mean and this tells us there is little diversity in the data set.
1. =mode(c:c)=25
2. Not skewed because the mean and median are close
3. minimun= 3, max.= 87, range= 84
4. 18 is Interquartile Range. The range of Q1 and Q3 are outliers
5. standard deviation 13.13667
1. After calculating all the data through excel and using formula =mode(c2:c1094). The mode came out to 25.
2. The graph and data is not skewed because the mean (36.81) and the median (35) are too close in value.
3. Min: 3
Max: 87
Range: 84
4. Interquartile range: 18 because the 75th percentile in age is 45 and the 25th percentile in age is 27 and when you subtract the two you get 18. This means that numbers greater than 45 and numbers less than 27 would be outliers.
5. The standard deviation is 13.13061. This means that since the standard deviation is low in value, the data is less spread out and closer to the average.
1. The most common occurring age of victims is 25 followed by 31 and 38.
2. No, the age variable is not skewed because there is no big difference between the average age (36.8) and the median age (35).
3. Min=3, max=87, range =84.
4. Interquartile range=18. This tells us the range of the center 50%, which is the median of the age distribution.
5. Standard Deviation=13.1. Since the standard deviation of this data has lower value, it suggests that there is less diversity which means the data is closer to the average (36.8).
1. =mode(c:c)=25
2. Not skewed because the mean/median are too close.
3. 3, 87, 84
4. 18, this method eliminates outliers to get a more general sense of the distribution of data without extreme values.
5. 13.13667, the data is less widely dispersed, data scores tend to be closer to the mode.
1) There are multiple modes; =mode(c:c) = 25
2)Median = 35 Mean =36.81181; No skew numbers are too close
3)Min= 3, Max= 87 Range =84
4)Percentile 75th= 45 Percentile 25th =27 IQR= 18, numbers outside Q1 and Q3 are outliers.
5)Standard Deviation = 13.13667 Standard deviation is close to the average.
1. 25
2. it is a negative skewed because the mean and median are close
3. mini: 3 Max: 87 range: 84
4. 18 because it is the middle distribution between Q1 and Q2 which is 27 and 45
5. the standard is 13.14 -> stdev (C:C)
1. They are many modes. I used the formula which is mode=(C2:C1094) in the excel sheet and I got 25.
2. No, the age variable is not skewed because the mean is 36.81181 and the median is 35. Also, because they are close to each other.
3. Minimum= 3
Maximum= 87
Range= 84
4. The Interquartile Range is 18. This means that the Interquartile Range is between Quartile 1 and Quartile 3 of the numbers 27 and 45.
5. The standard deviation of the variable is 13.13667. To interpret the standard deviation is that the numbers in the data are showing how its close to the mean. Also, because it shows how the standard deviation has a small number.
1. From the chart, I can tell that there are multiple modes, however when I put the formula
=mode (C2:C1094) into the excel file it gave me a final answer of 25.
2. The age variable is not skewed because the mean=36.81181 and the median=35, and both of these values are close in range.
3. Min=3
Max=87
Range=84
4. The interquartile range is 18, and it tells us that the IQR is the difference of Q1 and Q3 numbers which is 27 and 45, meaning all the numbers before 27 and after 45 are outliers.
5. The standard deviation is 13.13667 which means that the values in the data set are closer to the mean because the standard deviation is low.