Question

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be negative (-500). How would the value of the standard deviation change?

Answer 1

`\bar X_B = (\sum_(i=1)^5 X_i)/(5) =(500+200+250+275+300)/(5)=(1525)/(5)=305`

`s_B = \sqrt{(\sum_(i=1)^5 (X_i-\bar X)^2)/(n-1)}=\sqrt{((500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2))/(5-1)} = 115.108`

`\bar X_A = (\sum_(i=1)^5 X_i)/(5) =(-500+200+250+275+300)/(5)=(525)/(5)=105`

`s_A = \sqrt{(\sum_(i=1)^5 (X_i-\bar X)^2)/(n-1)}=\sqrt{((-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2))/(5-1)} = 340.221`

The absolute difference is:

`Abs = |340.221-115.108|= 225.113`

If we find the % of change respect the before case we have this:

`\% Change = (|340.221-115.108|)/(115.108) *100 = 195.57\%`

So then is a big change.

Explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

`\bar X_B = (\sum_(i=1)^5 X_i)/(5) =(500+200+250+275+300)/(5)=(1525)/(5)=305`

And the sample deviation with the following formula:

`s_B = \sqrt{(\sum_(i=1)^5 (X_i-\bar X)^2)/(n-1)}=\sqrt{((500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2))/(5-1)} = 115.108`

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

`\bar X_A = (\sum_(i=1)^5 X_i)/(5) =(-500+200+250+275+300)/(5)=(525)/(5)=105`

And the sample deviation with the following formula:

`s_A = \sqrt{(\sum_(i=1)^5 (X_i-\bar X)^2)/(n-1)}=\sqrt{((-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2))/(5-1)} = 340.221`

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

`Abs = |340.221-115.108|= 225.113`

If we find the % of change respect the before case we have this:

`\% Change = (|340.221-115.108|)/(115.108) *100 = 195.57\%`

So then is a big change.