Question

A veterinarian’s office sees, on the average, three patients per hour. Find the probability that the veterinarian’s office sees:

(A) at most three patients in one hour.

(B) at least three patients in one hour.

(C) five or more patients in one hour.

(D) less than 10 patients in four hours.

is this a binomial, geometric, hypergeometric, Poisson, normal, or a standard normal problem? I do not know how to tell the difference. Also if possible can the equations also be written with the work? I learn best when I see the equation right next to the work. Helps me remember. Thank you so much

Answer 1

Since here, only the rate at which veterinarian’s office sees patient is give so this is a poisson problem

Let X be the number of patients veterinarian’s office sees in one hour

X~ Poisson( $\lambda$ = 3 /hr)

a) P( at most three patients in one hour)

= P( X <=3)

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

= $\frac{e^{-3}3^0}{0!}$ + $\frac{e^{-3}3^1}{1!}$ +
e-332 2!
+ $\frac{e^{-3}3^3}{3!}$ ; ( pdf of poisson dist, f(x) =$\frac{e^{-\lambda}\lambda^x}{x!}$ )

=0.04979 + 0.14936 + 0.22404 + 0.22404

=0.64723

b) P( at least three patients in one hour)

= P( X >=3)

= 1- P(X<3)

= 1- [ P(X=0) + P(X=1) + P(X=2) ]

= 1- [  $\frac{e^{-3}3^0}{0!}$ + $\frac{e^{-3}3^1}{1!}$ + ]

e-332 2!

= 1- [0.04979 + 0.14936 + 0.22404 ]

= 1- 0.42319

=0.57681

c) P( five or more patients in one hour)

= P( X>=5)

=1- P(X<=4)

= 1- 0.81526 ; ( P(X<=4 ) can be directly seen from the poisson table OR can be calculated as above)

=0.18474

d) Now, the rate for four hours is ( 3*4) = 12 patients / 4hr

Now,  $\lambda$= 12/four hours

P(   less than 10 patients in four hours. )

= P( X < 10)

= P( X<=9) ; { same can be directly seen from the table or can be calculated in the above form}

= 0.24239