Question 834023: Exam scores are normally distributed with a mean of 81 and a standard deviation of 9.
a) What is the minimum score one must have to be in the top 4% of the students taking the exam?
b) What percentage of scores are below 90?
c) What percentage of scores are above 45?
d) If 30 people take this exam, how many would you expect to get above a 93%?
e) If you have a test score in the 45th percentile, what is its z-score?
f) What is the z-score and percentile for an exam score 1.2 standard deviations below the mean?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Exam scores are normally distributed with a mean of 81
and a standard deviation of 9.
a) What is the minimum score one must have to be in the top 4% of the students taking the exam?
Find the z-score with a right tail of 4%:
invNorm(0.96) = 1.7507
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Find the corresponding score: 81+1.7507*9 = 96.76
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b) What percentage of scores are below 90?
z(90) = (90-81)/9 = 1
P(x < 90) = P(z < 1) = 0.8413
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c) What percentage of scores are above 45?
Find z(45) then find the Probability z is above that z value.
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d) If 30 people take this exam, how many would you expect to get above a 93%?
Ans:0.07*30 = 2.1
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e) If you have a test score in the 45th percentile, what is its z-score?
invNorm(0.45) = -0.1257
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f) What is the z-score and percentile for an exam score 1.2 standard deviations below the mean?
z = -1.2
score = 81-1.2*9 = 70.2
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Cheers,
Stan H.
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