A sample of final exam scores is normally distributed with a mean equal to 23 and a variance equal to 16. Part (a) What percentage of scores are between 19 and 27? (Round your answer to two decimal places.)

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%