Question

If telephone purchase orders are Poisson distributed with an average of 3 orders per minute, answer the following:

1. probability at most 3 in a 30 second interval

2. probability of at least 4 in a 45 second interval

3. probability of 1, 2, or 3, in a 60 second interval

4. probability that of 5 different 45 second intervals 2 or more of the intervals would entail at least 4 orders each

5. standard deviation of the expected number of failures in a 60 second interval

Answer 1

1.

Given, = 3 order per minute = 3/60 order per second = (1/20)  order per second

For Poisson distribution in given time interval t, P(X = x) = exp(-t) (t)^x /x!

Probability at most 3 in a 30 second interval = P(X 3)

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

= exp(-30/20) (30/20)⁰/0! + exp(-30/20) (30/20)¹/1! + exp(-30/20) (30/20)²/2! + exp(-30/20) (30/20)³/3!

= exp(-1.5) + exp(-1.5) * 1.5 + exp(-1.5) * (1.5)² / 2 + exp(-1.5) * (1.5)³ /6

= 0.2231 + 0.3347 + 0.2510 + 0.1255

= 0.9343

2.

Probability of at least 4 in a 45 second interval = P(X 4) = 1 - P(X < 4)

= 1 - [P(X = 0) +  P(X = 1) +  P(X = 2) +  P(X = 3)]

= 1 - [ exp(-45/20) (45/20)⁰/0! + exp(-45/20) (45/20)¹/1! + exp(-45/20) (45/20)²/2! + exp(-45/20) (45/20)³/3! ]

= 1 - [exp(-2.25) + exp(-2.25) * 2.25 + exp(-2.25) * (2.25)² / 2 + exp(-2.25) * (2.25)³ /6]

= 1 - [0.1054 + 0.2371 + 0.2668 + 0.2]

= 1 - 0.8093

= 0.1907

3.

Probability of 1 in a 60 second interval = P(X = 1)

=  exp(-60/20) (60/20)¹/1!

= exp(-3) * 3 = 0.1494

Probability of 2 in a 60 second interval = P(X = 1)

=  exp(-60/20) (60/20)²/2!

= exp(-3) * 3² / 2 = 0.2240

Probability of 3 in a 60 second interval = P(X = 1)

=  exp(-60/20) (60/20)³/3!

= exp(-3) * 3³ / 6 = 0.2240

4.

From 2, Probability of at least 4 in a 45 second interval = 0.1907

Using Binomial probability, probability that of 5 different 45 second intervals 2 or more of the intervals would entail at least 4 orders each = P(Y 2) = 1 - P(Y < 2) {where Y ~ Binomial(5, 0.1907)}

= 1 - [P(Y = 0) + P(Y = 1)]

= 1 - [ ⁵C₀ * 0.1907⁰ * (1 - 0.1907)^5-0 +  ⁵C₁ * 0.1907¹ * (1 - 0.1907)^5-1 ]

= 1 - [0.3472 + 0.4090]

= 0.2438

5.

We know that the variance of Poisson distrinution is t

For 60 second interval, t = 60 * (1/20) = 3

So, variance of expected number of failures = 3

Standard deviation of the expected number of failures = = 1.732