After completing a study, orlando international airport managers concluded that the time needed to get passengers loaded on an airplane is normally distributed with a mean equal to 22.5 minutes and a standard deviation equal to 5.25 minutes. recently one airplane required 33 minutes to get passengers on board and ready for takeoff. find the probability that a flight will take 33 or more minutes to get passengers loaded.

Probability = 0.0228

Step-by-step explanation:

Let X be a random variable which is a measure of the time to get a passenger on board.

Mean(u) = 22.5mins

Standard deviation (s) = 5.25mins

X = 33mins

From Pr(X>33), starting from Pr(X = 33)

For normal distribution, Z = 7(X - u) /s

Z= (33 -22.5) /5.25

= 10.5/5.25

= 2

From the normal distribution table, 2 = 0.4772

Recall that

If Z is positive,

Pr(x>a) = 0.5 - table value

Pr(x<a) = 0.5 + table value

Therefore;

Pr(X>33) = 0.5 - 04772

= 0.0228

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40