Hospital waiting room times are normally distributed with a mean of 38.12 minutes and a standard deviation of 8.63 minutes. What is the shortest wait time that would still be in the worst 5% of wait times? Homework Help: 4VE. Determining values from normal distributions based on probabilities (Links to an external site.) (2:42) 4DC. Using normal distributions and probabilities to determine set values DOCX Group of answer choices 29.49 minutes 52.32 minutes 23.92 minutes 46.75 minutes

Step-by-step explanation:

waiting times follow a normal distribution with

Mean, \mu=38.12Mean,μ=38.12

Standard\ deviation,\sigma=8.63Standard deviation,σ=8.63

Longer waiting times are worse than shorter waiting times. Hence the worst 20% of wait times are wait times on the right tail of the distribution. The inferred level of confidence is 0.80.

The z value corresponding to the right tail probability of 0.2 is

Z=0.85Z=0.85

But

Z = \frac{x-\mu}{\sigma}Z=

σ

x−μ

​

x =Z*\sigma +\mux=Z∗σ+μ

=0.85 * 8.63 +38.12 =45.4555=0.85∗8.63+38.12=45.4555

the shortest wait time that would still be in the worst 20% of wait times is 45.4555 minutes

i need a picure

step-by-step explanation: