Answer: 0.03855
Step-by-step explanation:
Given :A population of skiers has a distribution of weights with mean 190 pounds and standard deviation 40 pounds.
Its maximum safe load is 10000 pounds.
Let X denotes the weight of 50 people.
As per given ,
Population mean weight of 50 people = [tex]\mu=50\times190=9500\text{ pounds}[/tex]
Standard deviation of 50 people [tex]=\sigma=40\sqrt{50}=40(7.07106781187)=282.84[/tex]
Then , the probability its maximum safe load will be exceeded =
[tex]P(X>10000)=P(\dfrac{X-\mu}{\sigma}>\dfrac{10000-9500}{282.84})\\\\=P(z>1.7671-8)\\\\=1-P(z\leq1.7678)\ \ \ \ [\because\ P(Z>z)=P(Z\leq z)]\\\\=1-0.96145\ \ \ [\text{ By p-value of table}]\\\\=0.03855[/tex]
Thus , the probability its maximum safe load will be exceeded = 0.03855