Americans spend an average of 3 hours per day online. If the standard deviation is 32 minutes, use Chebyshevs theorm to find the range (in minutes) which at least 93.75% of Americans spend online. Fine the z score for an online time of 4.75 hours per day

So then the range would be given by:

x =4.75*60= 285 min we got:

So this value means that 4.75 hr = 285 min is a value approximately 3.3 deviations above the mean of the distribution.

Step-by-step explanation:

For this case we have the following data given:

represent the mean

represent the deviation

The Chebysev theorem states that we have at least of the data lying withink k standard deviations of the mean.

So for this case we want 93.75% of the data or 0.9375 if we find the value of k we got:

So then the range would be given by:

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The z score is given by:

And if we replace for x =4.75*60= 285 min we got:

So this value means that 4.75 hr = 285 min is a value approximately 3.3 deviations above the mean of the distribution.

m∠a is 90°

m∠b is 40°