Answer :
To determine who is relatively more overweight in their age/body type group, we can calculate the z-scores for both Mark and Henry. The z-score measures how many standard deviations an individual’s weight is away from the mean weight for their group.
Mark's Calculation:
- Mean weight for Mark's group: [tex]\mu_M = 145 \text{ lbs}[/tex]
- Standard deviation for Mark's group: [tex]\sigma_M = 15 \text{ lbs}[/tex]
- Mark's weight: [tex]x_M = 178 \text{ lbs}[/tex]
The formula for the z-score is:
[tex]z_M = \frac{x_M - \mu_M}{\sigma_M} = \frac{178 - 145}{15} \approx 2.20[/tex]
Henry's Calculation:
- Mean weight for Henry's group: [tex]\mu_H = 165 \text{ lbs}[/tex]
- Standard deviation for Henry's group: [tex]\sigma_H = 20 \text{ lbs}[/tex]
- Henry's weight: [tex]x_H = 204 \text{ lbs}[/tex]
The formula for the z-score is:
[tex]z_H = \frac{x_H - \mu_H}{\sigma_H} = \frac{204 - 165}{20} = 1.95[/tex]
Since Mark has a z-score of approximately 2.20 and Henry has a z-score of 1.95, Mark is relatively more overweight compared to his age/body type group.
For the Commute Time Problem:
(a) Percentage of Students with Commute Time between 25.1 and 56.3 minutes
- Mean commute time: [tex]\mu = 40.7 \text{ minutes}[/tex]
- Standard deviation: [tex]\sigma = 8.5 \text{ minutes}[/tex]
The values 25.1 and 56.3 are 1.83 standard deviations below and above the mean respectively:
[tex]z = \frac{25.1 - 40.7}{8.5} \approx -1.83[/tex]
[tex]z = \frac{56.3 - 40.7}{8.5} \approx 1.83[/tex]
Using Chebyshev's inequality (since we do not assume a normal distribution), the minimum percentage of students within these bounds is:
[tex]1 - \frac{1}{(1.83)^2} \approx 0.703 \text{ or } 70.3\%[/tex]
(b) Commute Time Within 1.5 Standard Deviations of the Mean
According to Chebyshev's inequality:
[tex]1 - \frac{1}{(1.5)^2} = 1 - \frac{4}{9} = \frac{5}{9} \text{ or about } 55.56\%[/tex]
Therefore, at least 55.56% of students have commute times within 1.5 standard deviations of the mean.
