1 On an average Friday a waitress gets no tip from 5 custome

1. On an average Friday, a waitress gets no tip from 5 customers. Find the probability that she will get no tip from 7 customers this Friday.

2. During a typical football game, a coach can expect 3.2 injuries. Find the probability that the team will have at most 1 injury in this game. A coach can expect 3.2 injuries: = 3.2. Random Variable: The number of injuries the team has in this game. We are interested in P(x 1).

3. A small life insurance company has determined that on the average it receives 6 death claims per day. Find the probability that the company receives at least seven death claims on a randomly selected day.

Solution

Q1.
rate of tip is, gets no tip from 5 customers
Possion Distribution
PMF of P.D is = f ( k ) = e-k kx / x!
Wher
k = parameter of the distribution.
x = is the number of independent trials

P( X = 7 ) = e ^-5 * 5^7 / 7! = 0.1044

Q2.
a coach can expect 3.2 injuries in game
a)
P( X < = 1) = P(X=1) + P(X=0) +   
= e^-3.2 * 3.2 ^ 1 / 1! + e^-3.2 * 7 ^ 0 / 0! +
= 0.1712
b)
P( X = 1 ) = e ^-3.2 * 3.2^1 / 1! = 0.1304

Q3.
average it receives 6 death claims per day
P( X < 7) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0) +   
= e^-3.2 * 1 ^ 6 / 6! + e^-3.2 * ^ 5 / 5! + e^-3.2 * ^ 4 / 4! + e^-3.2 * ^ 3 / 3! + e^-3.2 * ^ 2 / 2! + e^-3.2 * ^ 1 / 1! + e^-3.2 * ^ 0 / 0! +
= 0.9554

P( X > = 7 ) = 1 - P (X < 7) = 0.0446