Answer:
I. 60%
II. 75.4 kg
Step-by-step explanation:
We will use the z-scores and the standard normal distribution to answer this questions.
We have a normal distribution with mean 69 kg and variance 25 kg^2 (therefore, standard deviation of 5 kg).
I. What percentage of adult male in Boston weigh more than 72 kg?
We calculate the z-score for 72 kg and then calculate the associated probability:
[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{72-69}{5}=\dfrac{3}{5}=0.6\\\\\\P(X>72)=P(z>0.6)=0.274[/tex]
II. What must an adult male weigh in order to be among the heaviest 10% of the population?
We have to calculate tha z-score that satisfies:
[tex]P(z>z^*)=0.1[/tex]
This happens for z=1.28 (see attachment).
Then, we can calculate the weight using this transformation:
[tex]X=\mu+z^*\cdot\sigma=69+1.28\cdot 5=69+6.4=75.4[/tex]
