Question

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 20% of wait times?

Answer 1

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

answer:

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

Answer 2

`x= 45.46`

Step-by-step explanation:

given,

mean of hospital waiting (μ) = 38.12 min

standard deviation (σ) = 8.63 min

worst waiting time = 20%

the are will be same as the = 100 - 20%

= 80%

we have to determine the z- value according to 80% or 0.80

using z-table

z-value = 0.85

now, using formula

`Z = (x-\mu)/(\sigma)`

`0.85 = (x-38.12)/(8.63)`

`x-38.12 = 0.85* {8.63}`

`x= 45.46`