Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
