Question

) Indian Airline is planning to change from a designated-seat allotment plan to open seating system. The Airline believes that 97% of its customers approve of the idea. A sample of 50 was interviewed. Find out the following-

•   The expected number of people who will like the change.

•   Standard deviation for the binomial distribution.

3)Arrivals to a bank ATM machine are distributed according to a poisson distribution with a mean equal to three per 15-minutes. Find out the following:

Determine the probability that in a given 15-minute segment no customer will arrive at the ATM machine

What is the probability that exactly three customers will arrive in a 30-minute segment?

Answer 1

Indian Airline is planning to change from a designated-seat allotment plan to open seating system. The Airline believes that 97% of its customers approve of the idea. A sample of 50 was interviewed. Find out the following-

•   The expected number of people who will like the change.

We are given n = 50, p = 0.97, q = 1 – p = 1 – 0.97 = 0.03

Expected number = n*p = 50*0.97 = 48.5

•   Standard deviation for the binomial distribution.

We are given n = 50, p = 0.97, q = 1 – p = 1 – 0.97 = 0.03

Standard deviation = sqrt(npq) = sqrt(50*0.97*0.03) = 1.20623

3)Arrivals to a bank ATM machine are distributed according to a poisson distribution with a mean equal to three per 15-minutes. Find out the following:

Determine the probability that in a given 15-minute segment no customer will arrive at the ATM machine

We are given λ = 3 for 15-minutes.

We have to find P(X=0)

P(X=x) = λ^x*exp(-λ)/x!

P(X=0) = 3^0*exp(-3)/0!

P(X=0) = 1*exp(-3)/1

P(X=0) = exp(-3)

P(X=0) = 0.049787

Required probability = = 0.049787

What is the probability that exactly three customers will arrive in a 30-minute segment?

We are given λ = 3*2=6 for 30-minutes.

We have to find P(X=3)

P(X=x) = λ^x*exp(-λ)/x!

P(X=3) = 6^3*exp(-6)/3!

P(X=0) = 216*exp(-6)/6

P(X=0) = 0.089235078

Required probability = 0.089235078