Answered

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The time spent waiting in the line is approximately normally distributed. The mean waiting time is 6
minutes and the standard deviation of the waiting time is 2
minutes. Find the probability that a person will wait for more than 8
minutes. Round your answer to four decimal places.

Sagot :

rvkacademic

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

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