Question

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

Answer 1

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

$\dfrac{1+2+3+4+5+6+7+8+9}{9}=\dfrac{45}{9}=5$

Median = $(\dfrac{n+1}{2})^{th}=\dfrac{9+1}{2}=5^{th}=5$

If 9 is replaced by 99,

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

So, mean would be

$\dfrac{1+2+3+4+5+6+7+8+99}{9}=\dfrac{135}{9}=15$

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.