If the outlier is excluded from the following data set, what happens to the mean? 22, 20, 23, 2, 21, 25, 31, 28
A. It does not change.
B. There is no outlier in this data set.
C. It decreases.
D. It increases.

Explanation:

Let's add up the values we're given to get

22 + 20 + 23 + 2 + 21 + 25 + 31 + 28 = 172

Then divide by 8, since there are 8 values in this set, leading to 172/8 = 21.5

The mean of the original set is 21.5

We'll come back to this later.

If you sort the values of {22, 20, 23, 2, 21, 25, 31, 28} you would get the list {2, 20, 21, 22, 23, 25, 28, 31}

Cross off the first and last number to reduce things to this smaller set {20, 21, 22, 23, 25, 28}

Repeat again, cross off the first and last number, and we now have {21, 22, 23, 25}

Do that one more time to be left with {22,23}

The median is the value at the midpoint of 22 and 23. Add these up and divide by two:  (22+23)/2 = 45/2 = 22.5

The median is 22.5

The items lower than the median are L = {2, 20, 21, 22} while the items in the upper set above the median are U = {23, 25, 28, 31}.

You should find the median of set L is 20.5 and the median of set U is 26.5

Therefore Q1 and Q3 are 20.5 and 26.5 respectively

Q1 = 20.5

Q3 = 26.5

Use these two values to get the interquartile range (IQR)

IQR = Q3 - Q1

IQR = 26.5 - 20.5

IQR = 6

Then we can compute the lower fence which I'll call m

lower fence = Q1 - 1.5*IQR

lower fence = 20.5 - 1.5*6

lower fence = 11.5

We can see the data value "2" is smaller than the lower fence 11.5; anything below the lower fence is considered an outlier.

All of the previous section was to find if we had an outlier, and identify what that outlier is. We found the single outlier is 2.

We could have probably done all of that by simply noticing that 2 is far from the main cluster of other values in the 20s; however, sometimes picking the outlier isn't that straightforward. So it's good practice to use the method described above.

Anyways, we'll remove the outlier "2" and we're left with this set {22,20,23,21,25,31,28}

Add up those values to get

22 + 20 + 23 + 21 + 25 + 31 + 28 = 170

Then divide that by 7, since we tossed out one value, and we get

170/7 = 24.2857 approximately

This is new mean.

To summarize:

We can see that the mean has increased. The very small outlier pulls on the mean to get smaller. Removing said outlier will then increase the mean so that it gravitates toward the main cluster. The mean is the center so if you remove the far outlier, then the center gravitates toward where the other values are located.

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841