anonymous
  • anonymous
The scores received by students completing a geology exam are normally distributed. If the average score is 122 and the standard deviation is 35, what percentage of students scored between 52 and 122? (A) 34.0% (B) 37.5% (C) 42.0% (D) 47.5% (E) 48.0%
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous

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Plasmataco
  • Plasmataco
im bad at these, only 13, but I can try.
anonymous
  • anonymous
:D
anonymous
  • anonymous
any help is better than none
Plasmataco
  • Plasmataco
dunno. sry
ybarrap
  • ybarrap
$$ P(52\le X\le 122)\\ =P(52-\mu\le X-\mu\le 122-\mu)\\ =P(\cfrac{52-\mu}{\sigma}\le \cfrac{X-\mu}{\sigma}\le \cfrac{122-\mu}{\sigma})\\ $$ Does this make any sense?
ybarrap
  • ybarrap
This is converting the probability statement into a z-score, a normal random variable with mean 0 and standard deviation 1. This allows us to use the z-table to figure out the probability.
ybarrap
  • ybarrap
$$ =P(\cfrac{52-\mu}{\sigma}\le \cfrac{X-\mu}{\sigma}\le \cfrac{122-\mu}{\sigma})\\ =P(-2\le \cfrac{X-\mu}{\sigma}\le 0)\\ =P(0\ge \cfrac{X-\mu}{\sigma}\le 2)\\ $$ The last equation follows from symmetry Do you see what we did here?
ybarrap
  • ybarrap
We now need to apply this - http://www.wolframalpha.com/input/?i=probability+normal+distribution&f1=-1&f=NormalProbabilities.z%5Cu005f-1&a=*FVarOpt-_**NormalProbabilities.l-.*NormalProbabilities.r--&a=*FVarOpt.2-_**-.***NormalProbabilities.mu--.**NormalProbabilities.sigma---.**NormalProbabilities.z---
ybarrap
  • ybarrap
This should be sufficient

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