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Last years freahman class at big state university totaled 5,305 students. Of those l, 1,258 received a merit scholarship to help offset tuition cost their freshman year (although the amount varied per student). The amount a student received was N($3,456, $478). If the coat of full tuition was $4,250 last year, what percentage of students who received a merit scholarship did not receive enough to cover full tutoring?

 Aug 7, 2022
 #1
avatar+2446 
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Wait... so each student received 3,456,478 dollars?

 

Are you sure your question is typed correctly?

 Aug 7, 2022
 #2
avatar+33151 
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I think N($3,456, $478) is meant to indicate a normal distribution with mean $3,456 and standard deviation $478 (or possibly a variance of $478, as both forms are common, and the poster hasn't specified which).

Alan  Aug 8, 2022
 #3
avatar+2446 
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Hi Alan, nice seeing you here!

 

The question is written in an ambiguous fashion, and it is unclear what was meant. 

BuilderBoi  Aug 8, 2022
 #4
avatar+33151 
+2

I suspect the following is what is required:

 

Alan  Aug 8, 2022
 #5
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BuilderBoi, if you just used a kitchen scale, it would have resolved the ambiguity by  indicating that N($3,456, $478) is in the form of \(\mathbb{N}(\mu, \sigma)\) in this case there is no need for pot-laced candy or mushroom tea (though it might help Probably not).

 

Recently, I saw a demonstration of very sophisticated kitchen scale at university. This scale connects to an AI-nerualnetwork. The scale’s sensors along with the AI identify the foods, caloric values, glycemic index, and other parameters. The AI also calculates the mean probable weight gain (with a standard deviation), based on the metabolic parameters of the consumer. It’s easy to program the parameters: the consumer need only to stand and then sit on the scale for 8 seconds, and then spit on a sensor. The demonstrated scale had a mass/weight limit of 450Kg –well above the weight of the largest tub-of-lard in attendance.      

 

During the demonstration, one of the spectators plopped a bag of pot-laced candy on the scale. The AI correctly identified the contents and caloric value, but indicated an 85% probability of weight gain at a mean of (28.7) times the maximum value for the calorie count.

 

The engineer-tech, who was demonstrating the scale, queried the AI for an explanation. The AI extrapolated the stats for \(\Delta9THC \) causing the munchies in a standard population; from this, a probability of weight gain is calculated beyond the caloric value of the weighed food.

 

It is amazing how innovative technology can create a kitchen scale that can construct and solve statistical problems.  You should get one as soon as they are on the market. LOL

 

 

GA

--. .- 

Guest Aug 10, 2022
edited by Guest  Aug 10, 2022

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