A sample of final exam scores is normally distributed with a mean equal to 23 and a variance equal to 16. Part (a) What percenta ge of scores are between 19 and 27? (Round your answer to two decimal places.)
1 answer:
Answer:
Let X = score of final exam.
X~normal(23, 16)
(a)
percentage of score between 19 and 27
= P(19 < X < 27)
= P((19-23) / sqrt(16) < Z < (27 - 23) / sqrt(16))
= P(-1 < Z < 1)
= 1 - P(Z <= -1) - P(Z >= 1)
= 1 - P(Z >= 1) - P(Z >= 1)
= 1 - 2*P(Z >= 1)
= 1 - 2(0.1587)
= 0.6826
= 68.26%
According to Normal Distribution Table
P(Z>=1) = 1 - P(Z<1) = 1-0.8413
So the final percentage is 68.26%
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