SOLUTION: Scores on a recent national statistics exam were normally distributed with a mean of 90 and a standard deviation of 5. 1.What is the probability that a randomly selected exam will

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Question 1141427: Scores on a recent national statistics exam were normally distributed with a mean of 90 and a standard deviation of 5.
1.What is the probability that a randomly selected exam will have a score of at least 85?
2.What percentage of exams will have scores between 89 and 92?
3.If the top 2.5% of test scores receive merit awards, what is the lowest score eligible for an award?
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
Use the Normal distribution table of z-scores
:
1. Probability(P) (X > 85) = 1 - P (X < 85)
:
z-score = (85 - 90)/5 = -1
:
P (X > 85) = 1 - 0.1587 = 0.8413
:
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the probability that a randomly selected exam will have a score of at least 85 is 0.8413
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:
2. P (89 < X < 92) = P (X < 92) - P (X < 89)
:
z-score(92) = (92 - 90)/5 = 0.40
:
z-score(89) = (89 - 90)/5 = -0.20
:
P (89 < X < 92) = 0.6554 - 0.4207 = 0.2347
:
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percentage with scores between 89 and 92 is 23.47 %
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3. 1 - 0.025 = 0.975
:
0.975 is associated with z-score of 1.96
:
(X - 90)/5 = 1.96
:
X - 90 = 9.8
:
X = 99.8
:
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the lowest score eligible for an award is 99.8
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