Question

Listed below are the gross amounts​ (in millions of​ dollars) earned in box office receipts for a recent movie. The amounts are listed in order for the first 14 days of the movies release. Find the​ range, variance, and standard deviation of the data set. If you invested in this​ movie, what characteristic of the data set would you care about​ most, and is it a measure of center or​ variation? 60 58 55 53 47 45 44 43 25 25 20 18 7 5

Answer 1

Answer:

$Range = 60-5=55$

In order to find the variance and deviation we need to find the mean with this formula:

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

And replacing we got:

$\bar X =36.07$

Now we can find the variance with the following formula:

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

And replacing we got:

$s^2=357.610$

And the standard deviation would be:

$s = \sqrt{357.610}= 18.911$

For this case since the range observed is large is better to use a measure of variation in order to check the spread of the values and take a decision useful

Step-by-step explanation:

For this case we have the following dataset:

60 58 55 53 47 45 44 43 25 25 20 18 7 5

We can find the range with this formula:

$Range = Max-Min$

And replacing we got:

$Range = 60-5=55$

In order to find the variance and deviation we need to find the mean with this formula:

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

And replacing we got:

$\bar X =36.07$

Now we can find the variance with the following formula:

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

And replacing we got:

$s^2=357.610$

And the standard deviation would be:

$s = \sqrt{357.610}= 18.911$

For this case since the range observed is large is better to use a measure of variation in order to check the spread of the values and take a decision useful