Respuesta :
Answer:
[tex] X_(r) >> X_(n) [/tex]
The mean for this case would increase since is defined as:
[tex] \bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
The interquartile range would not change since the definition for the IQR is [tex] IQR =Q_3 -Q_1[/tex] and the quartiles are the same.
The standard deviation would not remain the same since by definition is:
[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
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.
Step-by-step explanation:
For this case we assume that we have a random sample given [tex] X_(1), X_(2) ,..., X_(n) [/tex] and for each observation [tex] X_i \sim N(\mu, \sigma)[/tex] since the problem states that the data is approximately normal.
Let's assume that the largest value on this sample is [tex]X_(n)[/tex] and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:
[tex] X_(r) >> X_(n) [/tex]
The mean for this case would increase since is defined as:
[tex] \bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
The interquartile range would not change since the definition for the IQR is [tex] IQR =Q_3 -Q_1[/tex] and the quartiles are the same.
The standard deviation would not remain the same since by definition is:
[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
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.
