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)

= 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%