### MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1

Abstract curve, painted by Wassily Kandinsky

Data:
1. Using the list of 17 numbers at the top of the page, the median of this data, rounded to two decimal places, is: 2.8.

Raw Data: 7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5
Sort data to be:
-1.9, -1.5, -1.1, -0.1, 0.5, 1.2, 1.8, 2.1, 2.8, 3.2, 3.9, 4.1, 6.5, 7.0, 7.2, 13.0, and 18.0
Thus, median is: 2.8

Verify by R:
> x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)
> median (x)
[1] 2.8

2. If you find the median using the original method (paper and pencil), you have to arrange the values into numeric order (True/False): TRUE

Colorful houses, painted by Wassily Kandinsky

3. The interquartile range for this data is (round each value to 3 decimal places): 6.55
Lower quartile (Q1) = 1/4(n + 1) = 1/4(17+1) th value = 4.5th value = (-0.1+0.5)/2 = 0.2
Upper quartile(Q3)= 3/4 (n + 1) th value = 3/4(17+1)th value= 13.5th value= (6.5+7.0)/2=6.75

Thus, interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) = 6.75 – 0.2 = 6.55

Verify by R:
> IQR(x)
[1] 6
Hence, the result slightly different with doing by hand

4. The formula for calculating the interquartile range is:
(show the formula and a citation to the source that you used).
Interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) or IQR = Q3 – Q1

Reference
Stat Trek. (2017). Statistics and Probability Dictionary. Interquartile Range. Retrieved from http://stattrek.com/statistics/dictionary.aspx?definition=Interquartile%20range

Abstract, painted by Wassily Kandinsky

5. Using techniques that we have studied in this course, the upper and lower cutoff points (rounded to three decimal places) for identifying outliers (in the given data sample) are: 16.575 and -9.625
(this is not a request to show any outliers--just the cutoff points that would determine what constitutes an outlier).  You may round to three decimal places.

Upper cutoff = Q3 + (1.5 * IQR) = 6.75 + (1.5*6.55) = 16.575
Lower cutoff = Q1 – (1.5 * IQR) = 0.2 – (1.5*6.55) = 0.2 - 9.825 = -9.625

6. The summary() command shows a list of outliers, if there are any (True/False): FALSE
Verify by R:
> x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
-1.900 0.500 2.800 3.924 6.500 18.000
Thus, no list of outliers

Painted by Wassily Kandinsky

7. The list of outlier values is: 18
(if there are none, write "NA").
First we find upper and lower cutoff as at question 5.
Since we have 16.575 and -9.625 as upper and lower cutoff, thus 18 is the only value outside upper and lower cutoff

8. The standard deviation of the list of 17 numbers is (round to 3 decimal places):
5.2495 = 5.25
We find standard deviation by following formula:

SD= 5.2495 = 5.25
Verify by R:
> sd(x)
[1] 5.249468

# To be continued to part 2

1. Love those paintings!☺

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
Abstract curve, painted by Wassily Kandinsky

2. I studied this so many years ago!! Nice to remember!!
xoox

marisasclosetblog.com

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
Data:
1. Using the list of 17 numbers at the top of the page, the median of this data, rounded to two decimal places, is: 2.8.

Raw Data: 7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5
Sort data to be:
-1.9, -1.5, -1.1, -0.1, 0.5, 1.2, 1.8, 2.1, 2.8, 3.2, 3.9, 4.1, 6.5, 7.0, 7.2, 13.0, and 18.0
Thus, median is: 2.8

Verify by R:
> x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)
> median (x)
[1] 2.8

3. A lot of numbers, leave it them that knows.

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
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2. If you find the median using the original method (paper and pencil), you have to arrange the values into numeric order (True/False): TRUE

4. nice paintings

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
3. The interquartile range for this data is (round each value to 3 decimal places): 6.55

Lower quartile (Q1) = 1/4(n + 1) = 1/4(17+1) th value = 4.5th value = (-0.1+0.5)/2 = 0.2

Upper quartile(Q3)= 3/4 (n + 1) th value = 3/4(17+1)th value= 13.5th value= (6.5+7.0)/2=6.75

Thus, interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) = 6.75 – 0.2 = 6.55

Verify by R:

> IQR(x)

[1] 6

Hence, the result slightly different with doing by hand

5. I studied statistics at the university!
Don't Call Me Fashion Blogger
Bloglovin'

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-

6. I remember taking statistics for one of my undergraduate courses for psychology!

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
4. The formula for calculating the interquartile range is:

(show the formula and a citation to the source that you used).

Interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) or IQR = Q3 – Q1

Reference

Stat Trek. (2017). Statistics and Probability Dictionary. Interquartile Range. Retrieved from http://stattrek.com/statistics/dictionary.aspx?definition=Interquartile%20range

7. Good evening,
I was pleasantly surprised to find my favorite artist Wassily Kandinsky in this post. I always was fascinated by his paintings.
Wish you a nice week!

1. He is one of top world painters right now.

Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
5. Using techniques that we have studied in this course, the upper and lower cutoff points (rounded to three decimal places) for identifying outliers (in the given data sample) are: 16.575 and -9.625

(this is not a request to show any outliers--just the cutoff points that would determine what constitutes an outlier). You may round to three decimal places.

Upper cutoff = Q3 + (1.5 * IQR) = 6.75 + (1.5*6.55) = 16.575

Lower cutoff = Q1 – (1.5 * IQR) = 0.2 – (1.5*6.55) = 0.2 - 9.825 = -9.625

8. I probably don't like statistics but I love to know it. It sounds great. Thanks.

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
6. The summary() command shows a list of outliers, if there are any (True/False): FALSE

Verify by R:

> x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)

> summary(x)

Min. 1st Qu. Median Mean 3rd Qu. Max.

-1.900 0.500 2.800 3.924 6.500 18.000

Thus, no list of outliers

9. Bonjour,

Je dirai que j'aime tout simplement ces oeuvres !

Gros bisous 🌸

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
7. The list of outlier values is: 18

(if there are none, write "NA").

First we find upper and lower cutoff as at question 5.

Since we have 16.575 and -9.625 as upper and lower cutoff, thus 18 is the only value outside upper and lower cutoff

10. Great! Have a nice day

1. Thank you to visit:
MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
-
8. The standard deviation of the list of 17 numbers is (round to 3 decimal places):

5.2495 = 5.25

We find standard deviation by following formula:

SD= 5.2495 = 5.25

Verify by R:

> sd(x)

[1] 5.249468