Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
The probability that a randomly selected male college student gains 15 lb or more during their freshman year is 11.6%
What is Probability ?
Probability is defined as the likeliness of an event to happen.
Let X be a random variable that shows the term "freshman 15" that claims that students typically gain 15lb during their freshman year at college.
It is given that
X follows is a normal distribution with a mean of 2.1 lb (μ) and a standard deviation (σ) of 10.8 lb.
Population Mean (μ) = 2.1
Population Standard Deviation (σ) = 10.8
We need to compute Pr(X≥15). The corresponding z-value needed to be computed is:
[tex]\rm Z_{lower} = \dfrac{ X_1 -\mu }{\sigma}\\\\Z_{lower} = \dfrac{ 15-2.1 }{10.8}\\\\\\Z_{lower} = 1.19[/tex]
Then the probability is given as
[tex]\rm Pr(X \geq 16 ) = Pr (\dfrac{X -21}{10.8} \geq \dfrac{15-21}{10.8})\\\\= Pr (Z \geq \dfrac{15-2.1}{10.8}\\\\= Pr (Z\geq 1.19)\\\\ = 0.1162[/tex]
Pr(X≥15)=0.1162. (11.6%)
The probability that a randomly selected male college student gains 15 lb or more during their freshman year is 11.6%
To know more about Probability
https://brainly.com/question/11234923
#SPJ1
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
