Answer:
0.1587
Step-by-step explanation:
Using the notation
X actual wait time
[tex]\mu[/tex] mean wait time equals 6 in this case
[tex]\sigma[/tex] standard deviation equals 2 in this case
z standard normal value corresponding to [tex]\mu[/tex] and [tex]\sigma[/tex]
We can calculate the z value corresponding to the values above using the formula
[tex]z= \dfrac{X-\mu}{\sigma}[/tex]
In this case
[tex]z_8 = \dfrac{8-6}{2} = 1[/tex]
The question is asking for P(X > 8) which corresponds to P(z > 1)
In other words what is the probability that the wait time is greater than 1 standard deviation
using a calculator and/or standard normal tables we get
P(z ≤ 1) = 0.84134474059651
or
P(X ≤ 8) = 0.84134474059651
P(X > 8) = 1 - 0.84134474059651
= 0.15865525940349
Rounded to 4 decimal places
P(X>8) = 0.1587
Answer
The probability that a person will wait for more than 8 minutes = 0.1587
