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anonymous

  • one year ago

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%

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  1. anonymous
    • one year ago
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    @shinebrightlikeadimon @KyanTheDoodle

  2. anonymous
    • one year ago
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    @perl

  3. anonymous
    • one year ago
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    @Hero

  4. Plasmataco
    • one year ago
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    im bad at these, only 13, but I can try.

  5. anonymous
    • one year ago
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    :D

  6. anonymous
    • one year ago
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    any help is better than none

  7. Plasmataco
    • one year ago
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    dunno. sry

  8. ybarrap
    • one year ago
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    $$ P(52\le X\le 122)\\ =P(52-\mu\le X-\mu\le 122-\mu)\\ =P(\cfrac{52-\mu}{\sigma}\le \cfrac{X-\mu}{\sigma}\le \cfrac{122-\mu}{\sigma})\\ $$ Does this make any sense?

  9. ybarrap
    • one year ago
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    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.

  10. ybarrap
    • one year ago
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    $$ =P(\cfrac{52-\mu}{\sigma}\le \cfrac{X-\mu}{\sigma}\le \cfrac{122-\mu}{\sigma})\\ =P(-2\le \cfrac{X-\mu}{\sigma}\le 0)\\ =P(0\ge \cfrac{X-\mu}{\sigma}\le 2)\\ $$ The last equation follows from symmetry Do you see what we did here?

  11. ybarrap
    • one year ago
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    This should be sufficient

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