SOLUTION: Trouble with z scores. Assuming the scores for this test are normally distributed with mean = 85% and standard deviation of 5%. What is the probability of making greater than 9

Question 912830: Trouble with z scores.
Assuming the scores for this test are normally distributed with mean = 85% and standard deviation of 5%.
What is the probability of making greater than 92%?
What percentage of students will score 78% or below?
I use the formula, z = (value - mean)/deviation. my train of thought is to do (93-85)/5 + (94-85)/85 and so on but it's not working out for me.
Found 2 solutions by ewatrrr, stanbon:
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Find pertinent z-score and Use table/calculator etc find the 'cumulative probability.
P(x ≤ 92) = P(z < (92-85)/5) = P(z < 1.4) = .9192 0r 91.92%
P(x > 92) = 1 - P(x ≤ 92) = 8.08%
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Assuming the scores for this test are normally distributed with mean = 85% and standard deviation of 5%.
What is the probability of making greater than 92%?
z(92) = (92-85)/5 = 7/5
P(x > 92) = P(z > 7/5) = normalcdf(7/5,100) = 0.0808
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What percentage of students will score 78% or below?
z(78) = (78-85)/5 = -17/5
P(x <= 78) = P(z <= -17/5) = normalcdf(-100,-17/5) = 0.000337
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Cheers,
Stan H.

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