7. Scores on a recent national Mathematics exam were normally distributed with a mean of 82 and a standard deviation of 7.
A. What is the probability that a randomly selected exam score is less than 70
B. What is the probability that a randomly selected exam score is greater than 90?
C. If the top 2.5% of test scores receive Merit awards, what is the lowest score necessary to receive a merit award?
Solution :
Given that ,
A.
P(x < 70) = P[(x - ) / < (70 - 82 ) / 7]
= P(z < -1.71)
= 0.0436
Probability = 0.0436
B.
P(x > 90) = 1 - P(x < 90)
= 1 - P[(x - ) / < (90 - 82) / 7)
= 1 - P(z < 1.14)
= 1 - 0.8729
= 0.1271
Probability = 0.1271
C.
Using standard normal table ,
P(Z > z) = 2.5%
1 - P(Z < z) = 0.025
P(Z < z) = 1 - 0.025
P(Z < 1.96) = 0.975
z = 1.96
Using z-score formula,
x = z * +
x = 1.96 * 7 + 82 = 95.72
lowest score = 95.72
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