Question

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

Answer 1

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

This is a question of poisson distribution. Since this tells us about an average value within a fixed time period.

X: no. of patients in one hour

X ~ Poix)

N Poi
($1596219515809_blob.png$ = 3) per hour

= 3

X= E(X)

e-1 P(X = ) ..!

= $\frac{e^{-3}3^x}{x!}$

We then substitute 'x' values to find for X = x

(A) at most three patients in one hour.

At most means 3 or less than 3

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

= $\frac{e^{-3}3^0}{0!}+\frac{e^{-3}3^1}{1!}+\frac{e^{-3}3^2}{2!}+\frac{e^{-3}3^3}{3!}$

ANS: 0.64723

(B) at least three patients in one hour.

At least means 3 or more than 3

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

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

Ans: 0.5768

(C) five or more patients in one hour.

P( X > = 5) = 1 - P(X < 5)

=1 - [P(X =0) +...P( X = 4)]

ans: 0.18474

(D) less than 10 patients in four hours.

Here the time period is 4 hours,

So if there are 3 on avg in one hour then in 4 hours

3 * 4 = 12 patients on avg

$Y\sim$( 12) per 4 hours

P(Y = y) =$\frac{e^{-12}12^y}{y!}$

P( Y < 10) = P(Y = 0) + P( Y = 1) + .....P(Y = 9)

ans: 0.75761