Answered

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 :

Answer:

[tex] Range = 60-5=55[/tex]

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

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex] \bar X =36.07 [/tex]

Now we can find the variance with the following formula:

[tex]s^2 =\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And replacing we got:

[tex] s^2=357.610[/tex]

And the standard deviation would be:

[tex] s = \sqrt{357.610}= 18.911[/tex]

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:

[tex] Range = Max-Min[/tex]

And replacing we got:

[tex] Range = 60-5=55[/tex]

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

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex] \bar X =36.07 [/tex]

Now we can find the variance with the following formula:

[tex]s^2 =\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And replacing we got:

[tex] s^2=357.610[/tex]

And the standard deviation would be:

[tex] s = \sqrt{357.610}= 18.911[/tex]

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

Other Questions