A package delivery service claims that no more than 5 percen

A package delivery service claims that no more than 5 percent of all packages arrive at the address late. Assuming that the conditions for the binomial hold, if a sample of size 10 packages is randomly selected and the 5 percent rate holds, what is the probability that more than 2 packages will be delivered late? Cars arrive at a toll booth on an exit ramp on Route 408 in Florida at an average rate o f 3.4 cars every five minutes. Assume that the arrival of cars at this toll booth during the next five minutes? Kobe Bryant, a professional basketball player in the NBA, has made 84% of his free throws during his career with the Los Angeles Lakers. Calculate the probability that Bryant will make exactly three of his next five free throws. According to the National Work Intensification Survey, 42 % of Americans felt their workload last year had responsibility when compared to the start of the recession in December 2007.

Solution

a) The probability function for random variable X is,

f(X) = X / 6 , X = 1,2 or 3.

The expected value of X is,

E(X) = X * f(X) dX (X from 1,2 or 3)

= X * (X/6) dX

= 1/6 X² dX

= 1/6 [ X³ / 3 ] (X is from 1 to 3)

= 1/6 [ 3³/3 - 1³/3 ]

E(X) = 1.444

Given that a package delivery service claims that no more than 5 percent of all packages arrive at the address late.

n = number of packages = 10

p =5 % = 5 /100 = 0.05

Assuming that the condition for the binomial hold.

We know that the arriving distribution means Poison distribution with parameter µ.

The mean of the distribution is, µ = n*p = 10 *0.05 = 0.5

Let X be a random variable number of packages arriving late.

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

The probability mass function of X is,

P(X =x) = ( e^-µ * µ^x ) / x!

P(X = 0) = e^-0.5 * (0.5^0) / 0! = 0.6065

P(X = 1) = e^-0.5 * (0.5^1) / 1! = 0.3033

P(X = 2) = e^-0.5 (0.5^2) / 2! = 0.0758

P(X>2) = 1 - [0.6065+0.3033+0.0758] = 1 - 0.9856 = 0.0144

Hence the answer.