Write the probability mass function of Binomial distribution and Poisson distribution. When the Binomial distribution will tend to Poisson distribution. Six coins are tossed 6400 times. Using the Poisson distribution, find
the probability of getting six head 2 times (write formula only, not calculation)
Answers
Answer:
Step-by-step explanation:
The Poisson Distribution
Assume that a large Fortune 500 company has set up a hotline as part of a policy to eliminate sexual harassment among their employees and to protect themselves from future suits.) This hotline receives an average of 3 calls per day that deal with sexual harassment. Obviously some days have more calls, and some have fewer. We want to model the distribution of calls over the course of an extended period of time. We will assume that there is no seasonal variation in the number of calls. This is a situation that is ideal for illustrating the Poisson distribution. (The word is capitalized because the distribution is named after a 19th century French mathematician named Simeon-Denis Poisson.)
Before I continue, let me point out something important about the problem as I have stated it. I said that the hotline receives 3 calls per day. I did not say was that 3 out of 20 calls concerned sexual harassment, or anything similar. In other words, I have told you how many calls were about sexual harassment, but have told you nothing about some other category of calls. This will become important when we compare this distribution to the binomial distribution.
The important parameter, in fact the only parameter, of the Poisson distribution is μ, which represents the mean of the distribution. In our case, μ = 3, because I have said that the average number of calls per day is 3.
The distribution that we seek would tell us the probability of 0, 1, 2, 3, ... calls per day. This probability is given by the Poisson distribution as
PoissonDist (1K)
For example, the probability of 2 calls about harassment in a day can be calculated as
If we let x (the number of calls) take on all values between 0 and some arbitrarily high number, and if we substitute m = 3, we will obtain the following values for p:
x p
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15 .0498
.1494
.2240
.2240
.1680
.1008
.0504
.0216
.0081
.0027
.0008
.0002
.0001
.0000
.0000
.0000
In this example, once the values of x exceed about 10, the probabilities are so low that there is little point in calculating them. This distribution is plotted below.