Answer:
(D)109
Step-by-step explanation:
Mean = 450 seconds
Standard deviation = 50
First, we determine the probability that the expected response time is between 400 seconds and 500 seconds, P(400<x<500)
Using the Z-Score,
[tex]P(\frac{x-\mu}{\sigma} <x < \frac{x-\mu}{\sigma})\\=P(\frac{400-450}{50} <x < \frac{500-450}{50})\\=P(-1<x<1)[/tex]
From the Z-Score table
P(-1<x<1) = 0.68269
The probability that the expected response time is between 400 seconds and 500 seconds is 0.68269.
Since there are 160 Emergencies
Number whose expected time is between 400 seconds and 500 seconds
[tex]=160 X 0.68269 \approx 109[/tex]
Answer:
For those who had "Times for an ambulance to respond to a medical emergency in a certain town are normally distributed with a mean of 450 seconds and a standard deviation of 50 seconds.
Suppose there are 97 emergencies in that town.
In about how many emergencies are the response times expected between 400 seconds and 500 seconds?" The previous answers' explanation is really useful, and i solved it for my question, and the answer was 66!
Step-by-step explanation:
66!
