SOLUTION: The scores on an exam follow a normal distribution with mean 79 and standard deviation 6.5. What is the approximate probability that a randomly selected exam taker had a score bel

Question 1197272: The scores on an exam follow a normal distribution with mean 79 and standard deviation 6.5. What is the approximate probability that a randomly selected exam taker had a score below 85.5? Give your answer as a decimal with two digits after the decimal.

Found 2 solutions by Theo, ewatrrr:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean is 79.
standard deviation is 6.5
z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.
for this problem:
z = (85.5 - 79) / 6.5 = 1
probability of getting a z-score below 1 = area to the left of the z-score = .8413447404 = .84 rounded to 2 decimal places.
that's the same as probability of getting a raw score below 85.5.

Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
µ = 79 and σ =6.5
P(x < 85.5) ?
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P(x < 85.5) = normalcdf(smaller, larger, µ, σ) = normalcdf(-9999, 85.5, 79, 6.5 ) = .84
-9999 a place holder for the 'smaller number.
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