Question

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%

Answer 1

@shinebrightlikeadimon @KyanTheDoodle

Answer 2

@perl

Answer 3

@Hero

Answer 4

im bad at these, only 13, but I can try.

Answer 5

:D

Answer 6

any help is better than none

Answer 7

dunno. sry

Answer 8

$$
P(52\le X\le 122)\\
=P(52-\mu\le X-\mu\le 122-\mu)\\
=P(\cfrac{52-\mu}{\sigma}\le \cfrac{X-\mu}{\sigma}\le \cfrac{122-\mu}{\sigma})\\
$$
Does this make any sense?

Answer 9

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.

Answer 10

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

Answer 11

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---

Answer 12

This should be sufficient