Police response time to an emergency call is the difference between the times the call is first received by the dispatcher and the 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 responce time has a normal distribution with a mean of 8.4 minutes and a standard deviation of 1.7 minutes. for a randomly recieved emegency call, what is the probability that the response time will be:

a) between 5 and 10 min
b) less than 5 min
c) more than 10 min

a) 0.804

b) 0.023

c) 0.173

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.4 minutes

Standard Deviation, σ = 1.7 minutes

We are given that the distribution of response time is a bell shaped distribution that is a normal distribution.

Formula:

a) P(between 5 and 10 min)

b) P(less than 5 min)

P(x < 5)

Calculation the value from standard normal z table, we have,  

c) P(more than 10 min)

P(x > 10)

Calculation the value from standard normal z table, we have,  

(a) 0.80364

(b) 0.02275

(c) 0.17361

Step-by-step explanation:

We are given that the police response time has a normal distribution with a mean of 8.4 minutes and a standard deviation of 1.7 minutes i.e.;

Mean, = 8.4 minutes        and          Standard deviation, = 1.7 minutes

Also, normal distribution is given by;

Z score = ~ N(0,1)

Let X = the response time

(a) P( between 5 and 10 min ) = P(5 <= X <= 10) = P(X <= 10) - P(X < 5)

    P(X <= 10) = P( <= ) = P(Z <= 0.94) = 0.82639

    P(X < 5) = P( < ) = P(Z < -2) = 1 - P(Z < 2) = 1 - 0.97725 = 0.02275

Therefore, P( between 5 and 10 min ) = 0.82639 - 0.02275 = 0.80364 .

(b) P(X < 5 min) = P( < ) = P(Z < -2) = 1 - P(Z < 2) = 1 - 0.97725 =  0.02275 .

(c) P(X > 10) = 1 - P(X <= 10) = 1 - 0.82639 = 0.17361 .

he kicked it up? is this the answer ?

step-by-step explanation: