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%

the answer is 24

step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."