Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and th
e time a patrol car radios that it has arrived at the scene. Over a long period of time, it has been determined that the police response time has a normal distribution with a mean of 8.9 minutes and a standard deviation of 2.0 minutes. For a randomly received emergency call, find the following probabilities. (Round your answers to four decimal places.) a) between 5 and 10 min
b) less than 5 min
c) more than 10 min
Police response time to an emergency call is the difference between the time the call is first received and the time a patrol car radios it has arrived at the scene
Mean(u)= 8.9mins
Standard deviation (α) = 2.0mins
Let X be the random variable which is a measure of the time to get a response from the police.
We first solve for b and c
b) The response time less than 5 mins= Pr(X<5)
For normal distribution
Z= (X - u) / α
For X= 5
Z= (5 -8.9) / 2
Z= -3.9/2
Z= -1.95
From the normal distribution table, Z= -1.95 is 0.4744
Φ(z) = 0.4744
When Z is negative
Pr(X<a) = 0.5 - Φ(z)
Pr(X<5) = 0.5 - 0.4744
= 0.0256
c) The response time more than 10mins= Pr(X>10)
For normal distribution
Z= (X - u) / α
For X= 10
Z= (10 - 8.9) / 2
Z= 1.1/2
Z= 0.55
From the normal distribution table, Z= 0.55 is 0.2088
Φ(z) = 0.2088
When Z is positive
Pr(X>a) = 0.5 - Φ(z)
Pr(X>5) = 0.5 - 0.2088
= 0.2912
c) The response time between 5 and 10mins= Pr(5<X<10)
The bushes will be located 6 meters apart from each other.
Step-by-step explanation:
Given that at a walking path 78 mile long Colleen planted 14 bushes the same distance apart along the path, to determine how far away are the 1st 2 bushes along the path, the following calculation must be performed:
78/13 = X
6 = X
Thus, since the first bush is at meter 0, the remaining 13 bushes will be located 6 meters apart.
