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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

Answer :

Answer:

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:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(between 5 and 10 min)

[tex]P(5 \leq x \leq 10) = P(\displaystyle\frac{5 - 8.4}{1.7} \leq z \leq \displaystyle\frac{10-8.4}{1.7}) = P(-2 \leq z \leq 0.941)\\\\= P(z \leq 0.941) - P(z < -2)\\= 0.827 - 0.023 = 0.804 = 80.4\%[/tex]

[tex]P(5 \leq x \leq 10) = 80.4\%[/tex]

b) P(less than 5 min)

P(x < 5)

[tex]P( x< 5) = P( z > \displaystyle\frac{5 - 8.4}{1.7}) = P(z < -2)[/tex]

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

[tex]P(x < 5) = 0.023= 2.3\%[/tex]

c) P(more than 10 min)

P(x > 10)

[tex]P( x > 10) = P( z > \displaystyle\frac{10 - 8.4}{1.7}) = P(z > 0.9411)[/tex]

[tex]= 1 - P(z \leq 0.9411)[/tex]

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

[tex]P(x > 10) = 1 - 0.827= 0.173= 17.3\%[/tex]

Answer:

(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, [tex]\mu[/tex] = 8.4 minutes        and          Standard deviation, [tex]\sigma[/tex] = 1.7 minutes

Also, normal distribution is given by;

Z score = [tex]\frac{X -\mu}{\sigma}[/tex] ~ 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( [tex]\frac{X -\mu}{\sigma}[/tex] <= [tex]\frac{10-8.4}{1.7}[/tex] ) = P(Z <= 0.94) = 0.82639

    P(X < 5) = P( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{5-8.4}{1.7}[/tex] ) = 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( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{5-8.4}{1.7}[/tex] ) = 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 .

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