Question

Please I need help with this question!

Answer 1

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:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{72-69}{5}=\dfrac{3}{5}=0.6\\\\\\P(X>72)=P(z>0.6)=0.274$

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:

$P(z>z^*)=0.1$

This happens for z=1.28 (see attachment).

Then, we can calculate the weight using this transformation:

$X=\mu+z^*\cdot\sigma=69+1.28\cdot 5=69+6.4=75.4$