For the data in the following sample: 10, 6, 8, 6, 5

a. Find the mean and the standard deviation.
b. Now change the score of X=10 to X=0 , and find the new mean and standard deviation.
c. Describe how one extreme score influences the mean and standard devation.

a.

mean=7

standard deviation=2

b.

mean=5

standard deviation=3

c.

Due to one extreme score, our center of location changes and there comes more variability in the data measured through standard deviation.

Step-by-step explanation:

a.

Mean= sum of values/number of values

Mean=(10+6+8+6+5)/5

Mean=7

x     x-xbar  (x-xbar)²

10    3              9  

6    -1               1

8     1               1

6    -1               1

5    -2            4

sum(x-xbar)²=9+1+1+1+4=16

sum(x-xbar)²/n-1=16/4=4

Standard deviation=√4

Standard deviation=2

b.

Mean= sum of values/number of values

Mean=(0+6+8+6+5)/5

Mean=25/5

Mean=5

x     x-xbar  (x-xbar)²

0    -5              25  

6     1                1

8     3               9

6    1                 1

5    0               0

sum(x-xbar)²=25+1+9+1+0=36

sum(x-xbar)²/n-1=36/4=9

Standard deviation=√9

Standard deviation=3

c)

We can see that by adding one extreme value 0 our mean decreases from 7 to 5 and standard deviation increases from 2 to 3. This means that our center of location changes and there comes more variability in the data measured through standard deviation.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

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