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(20) 13. The number of people arriving at a doctor’s office is 4 per hour. Let...

(20) 13. The number of people arriving at a doctor’s office is 4 per hour. Let x represent the number of people arriving per hour.

(5) a. What is the probability that exactly 4 people arrive in one hour? Show how you calculated this.

(5) b. What is the probability that less than 2 people arrive in one hour? Show how you calculated this.

(5) c. What is the probability that less than 3 people arrive in one hour? Show how you calculated this.

(5) d. What is the probability that more than 3 people arrive in one hour? Show how you calculated this.

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Answer #1

Given the mean number of successes (μ) that occur in a specified region and actual number of successes (x) that result from the experiment,  the Poisson probability is given by

P(x; и) е р т!

Given

number of people arriving at a doctor’s office = \mu = 4 per hour

Let x represent the number of people arriving per hour.

a)

P(X=4)=\frac{e^{-4}*4^4}{4!}\approx 0.1954

The probability that exactly 4 people arrive in one hour is 0.1954

b)

P(X<2)=P(X=0)+P(X=1)

=\frac{e^{-4}*4^0}{0!}+\frac{e^{-4}*4^1}{1!}

\approx 0.0183 + 0.0733

= 0.0916

The probability that less than 2 people arrive in one hour is 0.0916

c)

P(X<3)=P(X=0)+P(X=1)+P(X=2)

=\frac{e^{-4}*4^0}{0!}+\frac{e^{-4}*4^1}{1!}+\frac{e^{-4}*4^2}{2!}

\approx 0.0183 + 0.0733 + 0.1465

= 0.2381

The probability that less than 3 people arrive in one hour is 0.2381

d)

P(X 3) = 1 - P(X <3

=1-(P(X< 3)+P(X=3))

=1-(0.2381+\frac{e^{-4}*4^3}{3!})

=1-(0.2381+0.1954)

=1-0.4335

= 0.5665

The probability that more than 3 people arrive in one hour is 0.5665

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