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3 years ago

When 'x' and 'y' are in direct proportion , then the relation between 'x' and 'y' is __________

Mathematics

1 answer:

Bess [88]2 years ago

5 0

Step-by-step explanation:

If two quantities x and y vary (change) together in such a manner that the ratio of their corresponding values remains constant, then x and y are said to be in direct proportion.

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Answer:

$X_(r) >> X_(n)$

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

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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:

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

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.

Step-by-step 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:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

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:

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

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.

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