Answer to Question #174272 in Statistics and Probability for Kwame Hay

Question

Question 3

a) State the Poisson Equation and explain its significance to an industry AP (6 marks)

3

b) An average of 10 cars per minute pass through a toll booth during rush hour. Using the Poisson distribution, find the probability that less than 6 cars pass through the toll booth during a randomly chosen minute and explain the significance of your

AN (6 marks)?

c) The probability distribution of lunch customers at a restaurant is given in Table 2, you are required to calculate expected number of lunch customers, the variance, and the standard deviation. Explain the significance of your answer in each case.

EV (7 marks)

Total Marks: 20

Table 2

Number of customers

Probability

100

110

118

120

125

0.2

0.3

0.2

0.2

0.1

Answer 1

a) The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

$P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$

where $k$ is the number of successes,

$\lambda$ is mean number of successes in the given time interval or region of space

The Poisson distribution can be applied to systems with a large number of possible events, each of which is rare.

Applications:

• birth defects and genetic mutations
• car accidents
• number of typing errors on a page
• failure of a machine in one month etc.

b) Using the Poisson distribution, the probability that less than 6 cars pass through the toll booth during a randomly chosen minute:

$P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+$

$+P(X=5)$

where $X$ is the number of cars pass through the toll booth during a randomly chosen minute.

We have an average of 10 cars per minute pass through a toll booth:

$\lambda=10$

Then:

$P(X=0)=e^{-10}$

$P(X=1)=10e^{-10}$

$P(X=2)=50e^{-10}$

$P(X=3)=500e^{-10}/3$

$P(X=4)=1250e^{-10}/3$

$P(X=5)=2500e^{-10}/3$

$P(X<6)=e^{-10}(61+4250/3)=0.0671=6.71\%$

c) Expected number of lunch customers is sum of products of number of customers and probability of this number:

$E(X)=\sum n_ip_i$

$E(X)=100\cdot0.2+110\cdot0.3+118\cdot0.2+120\cdot0.2+125\cdot0.1=113.1$

So, the average number of lunch customers at a restaurant is 113.1 customers.

Variance is the expectation of the squared deviation of a random variable (number of lunch customers) from its mean (average number of lunch customers). It measures how far a set of numbers is spread out from their average value.

$Var(X)=E[(X-E[X])^2]=E[X^2]-E[X]^2$

$Var(X)=\sum n^2_ip_i-E(X)^2$

$E[X^2]=100^2\cdot0.2+110^2\cdot0.3+118^2\cdot0.2+120^2\cdot0.2+125^2\cdot0.1=12857.3$

$Var(X)=12857.3-113.1^2=65.69$

 The standard deviation is a measure of the amount of variation or dispersion of a set of values.

$\sigma=\sqrt{Var(X)}=\sqrt{65.69}=8.10$