Americans spend an average of 5 hours per day online. If the standard deviation is 37 minutes, find the range in which at least 93.76% of the data will lie.

Question

Americans spend an average of 5 hours per day online. If the standard deviation is 37 minutes, find the range in which at least 93.76% of the data will lie.

 

 

Answer 1

Obtain a z=score table

http://www.utdallas.edu/dept/abp/zscoretable.pdf

 

0.9376 is not in the table. It lies half way between 0.9370 and 0.9382.

The corresponding z-scores are 1.53 and 1.54

The z-score lies half way between 1.53 and 1.54, or 1.535

 

z = (X - μ) / σ, where X = score, μ = mean, σ = standard deviation

 

1.535 = (X - 300) / 37                    changed 5 hours to 300 minutes

56.795 = X - 300

356.795 = X

 

356.795 minutes is 5.94658333 hours or 5 hours 56.795 minutes