Question

This problem illustrates the limit derivation of a Poisson distribution from Binomial distributions. Suppose an average of 66 arrivals occur during a 30 minute interval. To count arrivals, divide the 30 minute interval into nn sub-intervals. On the previous problem, you found the probability pp of one arrival during a single sub-interval for each n given. Now, compute the (estimated) probability that there will be, in fact, exactly 6 arrivals during a 30 minute interval, with each probability model:
a. Using Binomial with n=30, the chance of 6 arrivals is estimated as:__________
b. Using Binomial with n=60, the chance of 6 arrivals is estimated as:__________
c. Using Binomial with n=100n=100, the chance of 6 arrivals is estimated as:________
d. Using Poisson, the chance of 6 arrivals is estimated as:________

Required:
Use a probability calculator, and enter answers in decimal form, rounded to at least four places after the decimal.

Answer 1

Answer:

1. 0.1795

2. 0.1693

3. 0.1657

4. 0.1606

Step-by-step explanation:

This problem was solved using the binomial distribution

1.

N = 30

P = 6/30 = 0.2

Using binomial distribution

30C6 x 0.2⁶ x 0.8²⁴

= 593775 x 0.000064 x 0.004722

= 0.1795

2.

N = 60

P = 6/60 = 0.1

60C6 x 0.1⁶ x 0.9⁵⁴

= 50063860 x 0.000001 x 0.00338139

= 0.1693

3.

N = 100

100C6 x 0.06⁶ x 0.94⁹⁴

= 1192052400 x 0.000000046656 x 0.00297864

= 0.1657

4.

Poisson distribution was used here

e^-6(6)⁶/6!

= 0.1606