Respuesta :
Answer:
[tex]x= 45.46[/tex]
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
[tex]Z = \dfrac{x-\mu}{\sigma}[/tex]
[tex]0.85 = \dfrac{x-38.12}{8.63}[/tex]
[tex]x-38.12 = 0.85\times {8.63}[/tex]
[tex]x= 45.46[/tex]
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
