Question

Ambulance response time. Ambulance response time is measured as the time (in minutes) between the initial call to emergency medical services (EMS) and when the patient is reached by ambulance. Geographical Analysis (Vol. 41, 2009) investigated the characteristics of ambulance response time for EMS calls in Edmonton, Alberta. For a particular EMS station (call it Station A), ambulance response time is known to be normally distributed with m = 7.5 minutes and s = 2.5 minutes.

a. Regulations require that 90% of all emergency calls be reached in 9 minutes or less. Are the regulations met at EMS Station A? Explain.

b. A randomly selected EMS call in Edmonton has an ambulance response time of 2 minutes. Is it likely that this call was serviced by Station A? Explain

Short answer

⦁ 90% of the emergency calls can be reached in less than or equal to 9 minutes.

⦁ A randomly selected EMS call in Edmonton has an ambulance response time of 2 minutes.

Step-by-step solution

Given information

Ambulance response time is measured as the time (in minutes) between the initial call to emergency medical services (EMS) and when the patient is reached by ambulance.

For a particular EMS station ambulance response time is known to be normally distributed with m = 7.5 minutes and s = 2.5 minutes.

Verifying 90% of emergency calls can reach in less than or equal to 9 minutes.

⦁

Let X be the random variable which represents the ambulance response time with mean $\mu$and variance ${\sigma }^2$. That is $\mathrm{\mu }=7.5\min {\mathrm{\sigma }}^2=2.5\min$

Calculating Z-value of$X_1$

$$\begin{gathered} Z_1=\frac{X_1-\mu }{\sigma } \\ =\frac{9-7.5}{2.5} \\ =\frac{1.5}{2.5} \\ =0.6 \end{gathered}$$

If area A is 90% or more than 90%, then regulations are met at EMS station A.

Verifying a randomly selected eMS call in  Edmonton has an ambulance response time of 2 minutes.

⦁

Let t=2 minutes.

If probability of getting $X_2=2$ is less than 0.05, then it is unlikely that call was service by station A. That is

If area $A_2<0.05$

Therefore t is unlikely that call was serviced by station A.

Hence,

$$\begin{gathered} Z_1=\frac{X_2-\mu }{\sigma } \\ =\frac{2-7.5}{2.5} \\ {Z}_2=-2.2 \end{gathered}$$

From Z-table $Z_2=-2.2$, therefore $A_2=0.014$

Since $A_2\left(0.014\right)<0.05$, it is unlikely that the call was serviced by Station A.