Answer:
1.) 92.61%
2.) 97.58%
3.) 65.06%
Step-by-step explanation:
Mean populational weight (X) = 175 pounds
Standard deviation (σ) = 7.6 pounds
The z-score for any given student weight, X, is defined by:
[tex]z=\frac{X- \mu}{\sigma}[/tex]
The z-score can be converted to a percentile of the normal distribution through a z-score table.
1.
The z-score for X=186 pounds is:
[tex]z=\frac{186- 175}{7.6}\\z=1.447[/tex]
A z-score of 1.447 corresponds to the 92.61-th percentile.
Therefore, the probability that a male student weighs at most 186 pounds is:
[tex]P(X\leq 186) = 92.61\%[/tex]
2.
The z-score for X=160 pounds is:
[tex]z=\frac{160- 175}{7.6}\\z=-1.974[/tex]
A z-score of 1.447 corresponds to the 2.42-th percentile.
Therefore, the probability that a male student weighs at least 160 pounds is:
[tex]P(X\geq 160) = 100 - 2.42=97.58\%[/tex]
3.
The z-score for X=165 pounds is:
[tex]z=\frac{165- 175}{7.6}\\z=-1.316[/tex]
A z-score of -1.316 corresponds to the 9.41-th percentile.
The z-score for X=180 pounds is:
[tex]z=\frac{180- 175}{7.6}\\z=0.658[/tex]
A z-score of 0.658 corresponds to the 74.47-th percentile.
Therefore, the probability that a male student weighs between 165 and 180 pounds is:
[tex]P(165\leq X\leq 180) = 74.47-9.41=65.06\%[/tex]
