Question

Cheryl collected data for her mathematics project. She noted that the data set was approximately normal.

Select the statement that is always true if Cheryl replaced the maximum data value with a value that was an extremely high outlier.




Mean would increase.


Interquartile range would increase.


Standard deviation would remain the same.


The data set would remain approximately normal.

Answer 1

`X_(r) >> X_(n)`

The mean for this case would increase since is defined as:

`\bar X= (\sum_(i=1)^n X_i)/(n)`

The interquartile range would not change since the definition for the IQR is
`IQR =Q_3 -Q_1` and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

`s = \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

Explanation:

For this case we assume that we have a random sample given
`X_(1), X_(2) ,..., X_(n)` and for each observation
`X_i \sim N(\mu, \sigma)` since the problem states that the data is approximately normal.

Let's assume that the largest value on this sample is
`X_(n)` and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:

`X_(r) >> X_(n)`

The mean for this case would increase since is defined as:

`\bar X= (\sum_(i=1)^n X_i)/(n)`

The interquartile range would not change since the definition for the IQR is
`IQR =Q_3 -Q_1` and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

`s = \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.