add 2.8 and 11.6 and multiply 8 your answer will be 115.2 but when you subtract it with the four you will get the same thing when you calculate the regular problem.I hope this help

6 0

3 years ago

The term "freshman 15" refers to the claim that college students typically gain 15lbs during freshman year at college. Assume th

Viefleur [7K]

The probability that a randomly selected male college student gains 15 lb or more during their freshman year is 11.6%

<h3>What is Probability ?</h3>

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:

$\rm Z_{lower} = \dfrac{ X_1 -\mu }{\sigma}\\\\Z_{lower} = \dfrac{ 15-2.1 }{10.8}\\\\\\Z_{lower} = 1.19$

Then the probability is given as

$\rm Pr(X \geq 16 ) = Pr (\dfrac{X -21}{10.8} \geq \dfrac{15-21}{10.8})\\\\= Pr (Z \geq \dfrac{15-2.1}{10.8}\\\\= Pr (Z\geq 1.19)\\\\ = 0.1162$

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

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3 0

2 years ago