Abstract
curve, painted by Wassily
Kandinsky

**Tasks**

**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**.

Steps
to find the answer:

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**

Step
to find answers:

Please
see my answer at question 1 above

Colorful
houses, painted by Wassily
Kandinsky

3.
The interquartile range for this data is (round each value to 3
decimal places): **6.55**

Step
to find answers:

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

Upper
quartile(Q3)= 3/4 (n + 1) th value = 3/4(17+1)th value= 13.5^{th}
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).

Answer:

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.

Steps
to find answer:

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").

Steps
to find answer:

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 **

Step
to find answer:

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**