Respuesta :
Answer:
[tex] 0.9375 = 1- \frac{1}{k^2}[/tex]
[tex] \frac{1}{k^2} = 1-0.9375=0.0625[/tex]
[tex] k = \sqrt{\frac{1}{0.0625}}= 4[/tex]
So then the range would be given by:
[tex] Lower = \mu -4 \sigma = 180 -4*32=52 min[/tex]
[tex] Upper = \mu +4 \sigma = 180 +4*32=308min[/tex]
x =4.75*60= 285 min we got:
[tex] z = \frac{285-180}{32}=3.281[/tex]
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:
[tex] \mu = 3*60 = 180 min[/tex] represent the mean
[tex] \sigma= 32min[/tex] represent the deviation
The Chebysev theorem states that we have at least [tex] 1- \frac{1}{k^2}[/tex] 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:
[tex] 0.9375 = 1- \frac{1}{k^2}[/tex]
[tex] \frac{1}{k^2} = 1-0.9375=0.0625[/tex]
[tex] k = \sqrt{\frac{1}{0.0625}}= 4[/tex]
So then the range would be given by:
[tex] Lower = \mu -4 \sigma = 180 -4*32=52 min[/tex]
[tex] Upper = \mu +4 \sigma = 180 +4*32=308min[/tex]
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:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we replace for x =4.75*60= 285 min we got:
[tex] z = \frac{285-180}{32}=3.281[/tex]
So this value means that 4.75 hr = 285 min is a value approximately 3.3 deviations above the mean of the distribution.
