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 288 emergencies in that town. in about how many emergencies are the response times expected between 400 seconds and 500 seconds?

Question

Collin Noble

Mathematics

7 August, 09:15

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 288 emergencies in that town. in about how many emergencies are the response times expected between 400 seconds and 500 seconds?

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Answers (1)

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Jeramiah Mendoza · Answer 1

The number of emergencies between 400 sec and 500 sec will be determined as follows:

z = (x-μ) / σ

but

n=288

thus

when x=400

z = (400-450) / 50=-1

P (z<-1) = 0.4602

When x=500

z = (500-450) / 50=1

P (z<1) = 0.5398

Hence:

P (400
=0.5398-0.4602

=0.0796

thus the number of cases will be:

0.0796*288

=22.9~23 emergency cases