1. Consider the data set 1,2,3,4,5,6,7,8,9.

a. Obtain the mean and median of the data.

b. Replace the 9 in the data set by 99 and again compute the

mean and median. Decide which measure of center works bet-

ter here, and explain your answer.

c. For the data set in part (b), the mean is neither central nor

typical for the data. The lack of what property of the mean

accounts for this result?

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windyyork windyyork · Answer 1 · 2020-04-06T11:20:18+02:00

Answer: a) Mean = 5, Median = 5

b) Mean = 15, Median = 5

c) Due to presence of outlier i.e. 99.

Step-by-step explanation:

Since we have given that

1,2,3,4,5,6,7,8,9

Here, n = 9 which is odd

So, Mean would be

[tex]\dfrac{1+2+3+4+5+6+7+8+9}{9}=\dfrac{45}{9}=5[/tex]

Median = [tex](\dfrac{n+1}{2})^{th}=\dfrac{9+1}{2}=5^{th}=5[/tex]

If 9 is replaced by 99,

1,2,3,4,5,6,7,8,99

So, mean would be

[tex]\dfrac{1+2+3+4+5+6+7+8+99}{9}=\dfrac{135}{9}=15[/tex]

Median would be same as before i.e. 5

The mean is neither central nor typical for the data due to outlier i.e. 99

Hence, a) Mean = 5, Median = 5

b) Mean = 15, Median = 5

c) Due to presence of outlier i.e. 99.